BASIC LINEAR ALGEBRA - Kansas State Universitynagy/lin-alg/notes.pdfCHAPTER 1 Vector spaces and linear maps In this chapter we introduce the basic algebraic notions of vector spaces

BASIC LINEAR ALGEBRA

AN EXERCISE APPROACH

Gabriel Nagy

Kansas State University

c©Gabriel Nagy

CHAPTER 1

Vector spaces and linear maps

In this chapter we introduce the basic algebraic notions of vector spaces andlinear maps.

1. Vector spaces

Suppose k is a field. (Although the theory works for arbitrary fields, we willeventually focus only on the choices k = R, C.)

Definition. A k-vector space is an abelian group (V,+), equipped with anexternal operation1

k × V 3 (λ, v) 7−→ λv ∈ V,

called scalar multiplication, with the following properties:• λ · (v + w) = (λ · v) + (λ · w), for all λ ∈ k, v, w ∈ V .• (λ + µ) · v = (λv) + (µv), for all λ, µ ∈ k, v ∈ V .• (λ · µ)v = λ · (µ · v), for all λ, µ ∈ k, v ∈ V .• 1 · v = v, for all v ∈ V .

The elements of a vector space are sometimes called vectors.

Examples. The field k itself is a k-vector space, with its own multiplicationas scalar multiplication.

A trivial group (with one element) is always a k-vector space (with the onlypossible scalar multiplication).

1 Suppose V is a k-vector space. Prove that

0 · v = 0, for all v ∈ V.

(The zero in the left-hand side is the field k. The zero in the right-hand side is theneutral element in the abelian group V .)

Use the above fact to conclude that for any v ∈ V , the vector −v (the inverseof v in the abelian group V ) can also be described by

−v = (−1) · v.

2 Fix a field k, a non-empty set I, and a family (Vi)i∈I of k-vector spaces.Consider the product

∏i∈I Vk, equipped with the operations:

• (vi)i∈I + (wi)i∈I = (vi + wi)i∈I ;• λ · (vi)i∈I = (λvi)i∈I .

1 When convenient, the symbol · may be omitted.

1

2 1. VECTOR SPACES AND LINEAR MAPS

Prove that∏

i∈I Vk is a k-vector space. This structure is called the k-vector spacedirect product of the family (Vi)i∈I .

Definition. Suppose V is a k-vector space. A subset X ⊂ V is called ak-linear subspace, if

• Whenever x, y ∈ X, we have x + y ∈ X.• Whenever x, y ∈ Xand λ ∈ k, we have λx ∈ X.

3 If X is a k-linear subspace of the k-vector space V , then X itself is a k-vectorspace, when equipped with the operations “inherited” from V . Prove than anylinear subspace of V contains the zero vector 0 ∈ V

4 Let (Vi)i∈I be a family of k-vector spaces (indexed by a non-empty set I). Foran element v = (vi)i∈I ∈

∏i∈I Vi let us define the set

bvc = i ∈ I : vi 6= 0.

Prove that the set

v ∈∏i∈I

Vi : bvc is finite

is a linear subspace of∏

i∈I Vi. This space is called the k-vector space direct sumof the family (Vi)i∈I , and is denoted by

⊕i∈I Vi.

Definition. Suppose we have a family (Vi)i∈I of vector spaces. For a fixedindex j ∈ I, define the map εj : Vj →

⊕i∈I Vi as follows. For a vector v ∈ Vj we

construct εj(v) = (wi)i∈I , where

wi =

v if i = j0 if i 6= j.

We call the maps εj : Vj →⊕

i∈I Vi, j ∈ I, the standard inclusions. The maps

πj :∏i∈I

Vi 3 (vi)i∈I 7−→ vj ∈ Vj

are called the coordinate maps.

5 Let (Vi)i∈I be a family of vector spaces. Prove that the standard inclusionsεi, i ∈ I. are injective. In fact prove that πi εi = IdVi

. Prove that any elementv ∈

⊕i∈I Vi is given as

v =∑

i∈bvc

(εi πi)(v).

In other words, if v = (vi)i∈I , then v =∑

i∈bvc εi(vi).

6 Suppose (Xj)j∈J is a family of k-linear subspaces of V . Prove that the inter-section

⋂j∈J Xj is again a k-linear subspace of V .

1. VECTOR SPACES 3

Definition. Let V be a k-vector space, and let M ⊂ V be an arbitrary subsetof V . Consider the family

F = X : X k-linear subspace of V , and X ⊃M.

The setSpank(M) =

⋂X∈F

X,

which is a linear subspace of V by the preceding exercise, is called the k-linear spanof M in V .

Convention. Spank(∅) = 0.

Example. The linear span of a singleton is described as

Spank(v) = kv(= λv : λ ∈ k).

7 Prove that if M and N are subsets of a k-vector space V , with M ⊂ N , then wealso have the inclusion Spank(M) ⊂ Spank(N). Give an example where M ( N ,but their spans coincide.

8 Let V be a k-vector space, and M be a subset of V . For an element v ∈ V ,prove that the following are equivalent:

(i) v ∈ Spank(M);(ii) there exists an integer n ≥ 1, elements x1, . . . , xn ∈ M , and scalars

λ1, . . . , λn ∈ k such that2 v = λ1x1 + · · ·+ λnxn.

Hint: First prove that the set of elements satisfying property (ii) is a linear subspace. Second,

prove that the linear span of M contains all elements satisfying (ii).

Notation. Suppose V is a vector space, and A1, . . . , An are subsets of V . Wedefine

A1 + · · ·+ A2 = a1 + · · ·+ an : ak ∈ Ak, k = 1, . . . , n.

9 Let V be a k-vector space. Suppose A1, . . . , An are k-homogeneous, in thesense that for every k = 1, . . . , n we have the equality:

Ak = λx : λ ∈ k, x ∈ Ak.

Prove the equality

Span(A1 ∪ · · · ∪An) = Span(A1) + · · ·+ Span(An).

10 Let V be a k-vector space, and (Xj)j∈J be a family of linear subspaces of V .For an element v ∈ V , prove that the following are equivalent:

(i) v ∈ Spank

( ⋃j∈J Xj);

(ii) there exists an integer n ≥ 1 and x1, . . . , xn ∈⋃

j∈J Xj , such that v =x1 + · · ·+ xn.

2 From now on we will use the usual convention which gives the scalar multiplication prece-

dence over addition.


Comment. If X1, . . . , Xn are linear subspaces of the vector space V , then usingthe notation preceding Exercise ??, and the above result, we get

Span(X1 ∪ · · · ∪Xn) = X1 + · · ·+ Xn.

11 In general, a union of linear subspaces is not a linear subspace. Give anexample of two linear subspaces X1, X2 ⊂ R2, such that X1 ∪ X2 is not a linearsubspace.

12 Prove that the union of a directed family of linear subspaces is a linearsubspace. That is, if (Xj)j∈J is a family of k-linear subspaces of the k-vector spaceV , with the property

• For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk,then

⋃j∈J Xj is again a k-linear subspace of V .

Hint: Use the preceding exercise.

Definition. Suppose V is a k-vector space. A set M ⊂ V is said to be k-linearly independent, if (compare with Exercise ??) for every strict subset P ( M ,one has the strict inclusion Spank(P ) ( Spank(M).

13 Let M be a subset in the k-vector space V . Prove that the following areequivalent:

(i) M is linearly independent.(ii) If n ≥ 1 is an integer, if x1, . . . , xn ∈ M are different elements, and if

λ1, . . . , λn ∈ k satisfy

λ1x1 + · · ·+ λnxn = 0,

then λ1 = · · · = λn = 0.

Hint: To prove (i) ⇒ (ii) show that if one has a relation as in (ii), then one of the x’s can be

eliminated, without changing the linear Span.

14 Prove that a linearly independent set M cannot contain the zero element.Prove that a subset of a linearly independent set is again linearly independent.

15 Prove that the union of a directed family of linearly independent sets is againa linearly independent set. That is, if (Xj)j∈J is a family of k-linearly independentsubsets of the k-vector space V , with the property

• For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk,then

⋃j∈J Xj is again a k-linearly independent subset of V .

16 Suppose V is a vector space, and x1, x2, . . . is a sequence (finite or infinite)of different non-zero vectors in V . Prove that the following are equivalent:

(i) The set M = x1, x2, . . . is linearly independent.(ii) The sequence of susbspaces Wk = Span(x1, . . . , xk) is strictly increasing,

in the sense that we have strict inclusions

W1 ( W2 ( . . . .

1. VECTOR SPACES 5

Hint: The implication (i) ⇒ (ii) is clear from the definition.

Conversely, if M were not linearly independent, there exist scalars λ1, . . . , λn ∈ k such that

λ1x1 + · · · + λnxn = 0, and at least one of the λ’s non-zero. If we take k = maxj : 1 ≤ j ≤n and λj 6= 0, then we get xk ∈ Spanx1, . . . , xk−1, which proves that Wk = Wk−1.

Definition. Let V be a k-vector space. A subset B ⊂ V is called a k-linearbasis for V , if:

• Spank(B) = V ;• B is k-linearly independent.

17 Prove the following:

Theorem 1.1.1. Let V be a k-vector space, and let P ⊂ M ⊂ V be subsets,with P linearly independent, and Spank(M) = V . Then there exists a linear basisB such that P ⊂ B ⊂M .

Sketch of proof: Consider the set

B = B ⊂ V : B linearly independent, and P ⊂ B ⊂M,

equipped with the inclusion as the order relation. Use Zorn’s Lemma to prove that B has a

maximal element, and then prove that such a maximal element must span the whole space. (One

key step is the checking of the hypothesis of Zorn’s Lemma. Use Exercise 13??.)

18 Suppose B is a linear basis for the vector space V , and P and M are subsetsof V , such that one has strict inclusions P ( B ( M . Prove that P and M are nolonger bases (although P is linearly independent and Span(M) = V ).

19 Let W be a linear subspace of the vector space V , and let A be a linear basisfor W . Prove that there exists a linear basis B for V , with B ⊃ A.Hint: Use Theorem 1.1.1.

Definition. A vector space is said to be finite dimensional, if it has a finitebasis.

20 Prove the following

Lemma 1.1.1. (Exchange Lemma) Suppose A and B are linear bases for thevector space V . Prove that for every a ∈ A, there exists some b ∈ B, such that theset (A r a) ∪ b is again a linear basis.

Hint: Write a = β1b1 + · · · + βnbn, for some b1, . . . , bn ∈ B, and some β1, . . . , βn ∈ k r 0.At least one of the b’s does not belong to Span (Ar a). Choose this as the exchange for a.

We have the inclusion (Ar a) ∪ b ⊂ A ∪ b, with the first set linearly independent, but the

second one not. Using the fact that Span (A ∪ b) = V , Theorem 1.1.1 will force (Ar a)∪ bto be the basis.

21 Prove that any two linear bases, in a finite dimensional vector space, have thesame number of elements.Hint: Fix a finite basis A = a1, . . . , an, and let B be an arbitrary basis. Use the Exchange

Lemma 1.1.1 to construct inductively a sequence of elements b1, . . . , bn ∈ B, such that

• For every k = 1, . . . , n, the set b1, . . . , bk−1, ak, . . . , an is a linear basis.


Note that all the b’s must be different. We end up with a linear basis B0 = b1, . . . , bn, with

|B0| = n. Use Exercise ?? to conclude that B0 = B.

22* Prove the following generalization of the above result.

Theorem 1.1.2. Any two linear bases in a vector space have the same car-dinality. (In cardinal arithmetic two sets have the same cardinality if there is abijection between them.)

Sketch of proof: Suppose A and B are two bases for V . If either A or B is finite, we applythe previous exercise. So we can assume that both A and B are infinite.

For every a ∈ A we write it (uniquely) as a = β1b1 + · · · + βnbn, for some b1, . . . , bn ∈ B,

and some β1, . . . , βn ∈ k r 0, and we set PB(a) = b1, . . . , bn, so that PB(a) is the smallestsubset of P ⊂ B, with Span(P ) 3 a. If we denote by Fin(B) the set of all finite subsets of B, we

have now a map Π : A 3 a 7−→ PB(a) ∈ Fin(B). This map is not injective. However, for every

P ∈ Fin(B), we have Π−1(P ) ⊂ Span(P ), which means that Π−1(P ) is a linearly independent setin the finite dimensional vector space Span(P ). By Theorem 1.1.1, combined with the previousexercise, this forces each of the sets Π−1(P ), P ∈ Fin(B), to be finite.

Then A is a disjoint union of preimages A =⋃

P∈Fin(B) Π−1(P ), each of the sets in this

union being finite. Since Fin(B) is infinite, we get

Card(A) = Card( ⋃

P∈Fin(B)

Π−1(P ))≤ Card

(Fin(B)

)= Card(B).

By symmetry, we also have Card(B) ≤ Card(A), and we are done.

Definition. Given a k-vector space V , the above Theorem states that thecardinality of a linear basis for V is independent of the choice of the basis. This“number” is denoted by dimkV , and is called the dimension of V . In the finitedimensional case, the dimension is a non-negative integer. If V is the trivial (zero)space, we define its dimension to be zero.

23 Let n ≥ 1 be an integer. Define, for each j ∈ 1, . . . , n, the element ej =(δij)n

i=1 ∈ kn. (Here δij stands for the Kronecker symbol, defined to be 1, if i = j,and 0 if i 6= j.) Prove that the set e1, . . . ,en is a linear basis for the vector spacekn, therefore we have dim kn = n.

24 Generalize the above result, by proving that

dim( ⊕

i∈I

k)

= Card(I).

25 Suppose W is a linear subspace of the vector space V . Prove that dim W ≤dim V . (This means that if A is a linear basis for W , and B is a linear basis for V ,then Card(A) ≤ Card(B), which in turn means that there exists an injective mapf : A→ B.)Hint: Find another basis B′ for V , with B′ ⊃ A. Then use Theorem 1.1.2.

26 Suppose V is a vector space. Prove that the following are equivalent:(i) V is finite dimensional.(ii) Whenever W is a linear subspace of V , with dimW = dimV , it follows

that W = V .

1. VECTOR SPACES 7

(iii) Every infinite increasing sequence W1 ⊂W2 ⊂W3 ⊂ · · · ⊂ V is an infiniteof linear subspaces is stationary, in the sense that there exists some k ≥ 1,such that Wn = Wk, for all n ≥ k.

Definition. Suppose W = (Wj)j∈J is a family of non-zero linear subspacesof a vector space V . We say that W is linearly independent, if for every k ∈ J onehas

Wk ∩ Span( ⋃

j∈Jrk

Wj

)= 0.

27 Let V be a vector space, and X be a subset of V r 0. The following areequivalent:

(i) The set X is linearly independent (as a subset of V ).(ii) The family (kx)x∈X is a linearly independent family of linear subspaces

of V .

28 Let W = (Wi)i∈I be a linearly independent family of linear subspaces of V .Suppose X = (Xi)i∈I is another family of linear subspaces, such that Xi ⊂Wi, forall i ∈ I. Define J =

j ∈ I : Xj 6= 0

. Prove that XJ = (Xj)j∈J is also a

linearly independent family.

29 Suppose V is a vector space, and W1,W2, . . . is a sequence (finite or infinite)of non-zero linear subspaces of V . Prove that the following are equivalent:

(i) The sequence (W1,W2, . . . ) is linearly independent family.(ii) For every k we have

Wk ∩ (W1 + · · ·+ Wk) = 0.

Hint: Argue along the same lines of the hint given to exercise ??.

30 Let V be a vector space, and let W1 and W2 be two non-zero linear subspaces.Prove that (W1,W2) is a linearly independent pair of linear subspaces, if and onlyif W1 ∩W2 = 0.

31 Let W be a linear subspace of the vector space V . Prove that there exists alinear subspace X of V , such that W ∩X = 0, and W + X = V .Hint: Start with some linear basis A for W . Find a linear basis B for V , with B ⊃ A. Take

X = Span(B rA).

32 Suppose W = (Wj)j∈J is a family of linear subspaces of a vector space V .Prove that the following are equivalent

(i) W is linearly independent.(ii) If n ≥ 1 is an integer, if j1, . . . , jn ∈ J are different indices, and if w1 ∈

Wj1 , . . . , wn ∈ Wjnare elements such that w1 + · · · + wn = 0, it follows

that w1 = · · · = wn = 0.(iii) There exists a choice, for each j ∈ J , of a linear basis Bj for Wj , such

that(α) the sets (Bj)j∈J are mutually disjoint, i.e. for any j, k ∈ I with j 6= k,

we have Bj ∩Bk = ∅;(β) the set

⋃j∈J Bj is linearly independent.


(iii’) For any choice, for each j ∈ J , of a linear basis Bj for Wj , we have theproperties (α) and (β) above.

33 Let (Wj)j∈J be a linearly independent family of linear subspaces of V . If wechoose, for each j ∈ J , a linear basis Bj for Wj , then

⋃j∈J Bj is linear basis for

the vector space Span( ⋃

j∈J Wj

).

34 Let W1, . . . ,Wn be a finite linearly independent family of linear subspaces ofV . Prove that

dim (W1 + · · ·+ Wn) = dim W1 + · · ·+ dim Wn.

35 Let V be a finite dimensional vector space. Suppose W1, . . . ,Wn are linearsubspaces of V , with

• W1 + · · ·+ Wn = V ;• dim W1 + · · ·+ dim Wn.

Prove that (Wk)nk=1 is a linearly independent family.

Hint: For each k ∈ 1, . . . , n, let Bk be a linear basis for Wk. Put B = B1 ∪ · · · ∪Bn. On the

one hand, we have

(1) |B| ≤ |B1|+ · · ·+ |Bn| = dimW1 + · · ·+ dimWn = dimV.

On the other hand, we clearly have Span(B) = V , so we must have |B| ≥ dimV . This means that

we must have equality in (1), so in particular the sets B1, . . . , Bn are mutually disjoint, and their

union is a linear basis for V . Then the result follows from exercise ??

36 Let (Vi)i∈I be a family of vector spaces, and let εj : Vj →⊕

i∈I Vi, j ∈ I, bethe standard inclusions.

(i)(εj(Vj)

)j∈I

is a linearly independent family of linear subspaces of⊕

i∈I Vi.(ii) Suppose for each j ∈ I, we choose a linear basis Bj for Vj . Then εj(Bj)

is a linear basis for Vj .Using the fact that εj are injective, conclude that dim εj(Vj) = dim Vj , for all j ∈ I

As an application of the above, prove (use exercise ??) that the dimension of afinite direct sum is given as

dim( n⊕

i=1

Vi

)=

n∑i=1

dim Vi.

Definition. Suppose X is a k-linear subspace of the k-vector space V . Inparticular X is a subgroup of the abelian group (V,+) We can then define thequotient group V/X, which is again an abelian group. Formally, the quotient isdefined as the set of equivalence classes modulo X:

v ≡ w (mod X), if and only if v − w ∈ X.

The addition is defined as [v]X + [w]X = [v + w]X . (Here [v]X stands for theequivalence class of v.) We can also define the scalar multiplication by λ · [v]X =[λv]X . With these operations V/X becomes a vector space, called the quotientvector space. For a subset A ⊂ V , we denote by [A]X the set [a]X : a ∈ A ⊂ V/X.

1. VECTOR SPACES 9

37 Verify the statements made in the above definition.

Definitions. Suppose X is a k-linear subspace of the k-vector space V .• For a set M ⊂ V we define its k-linear span relative to X to be the linearsubspace

Spank(M ;X) = Spank(M ∪X) = X + Spank(M).

• A set P ⊂ V is said to be k-linearly independent relative to X, if themap P 3 p 7−→ [p]X ∈ V/X is injective, and the set [P ]X is linearlyindependent in the quotient vector space V/X.• A set B ⊂ V is called a k-linear X-basis for V ,

- Span(B;X) = V ;- B is linearly independent relative to X.

38 Let X be a linear subspace of the vector space V .A. If P ⊂ V is linearly independent relative to X, then P is linearly inde-

pendent.B. For a set P ⊂ V , the following are equivalent:

(i) P is linearly independent relative to X.(ii) If n ≥ 1 is an integer, if p1, . . . , pn ∈ P are different elements, and if

λ1, . . . , λn ∈ k satisfy

λ1p1 + · · ·+ λnpn ∈ X,

then λ1 = · · · = λn = 0.(iii) There exists a linear basis B for X, such that P ∪ B is linearly

independent.(iii’) If B is any linear basis for X, then P ∪B is linearly independent.(iv) P is linearly independent, and X ∩ Span(P ) = 0.

39 Let X and Y be linear subspaces of the vector space V , such that X ⊂ Y .A. Prove that for every set M ⊂ V , we have the inclusion Span(M ;X) ⊂

Span(M ;Y ).B. Prove that if a set P is linearly independent relative to Y , then P is also

linearly independent relative to X.

40 Let X be a linear subspace of the vector space V . For a set B ⊂ V prove thatthe following are equivalent:

(i) B is a linear X-basis for V .(ii) The map B 3 b 7−→ [b]X ∈ V/X is injective, and [B]X is a linear basis for

V/X.In this situation, prove the equality

dim V/X = Card(B).

41 Let X be a linear subspace of the vector space V , and let A be a linear basisfor X. Suppose B ⊂ V has the property A ∩ B = ∅. Prove that the following areequivalent:

(i) B is a linear X-basis for V .


(ii) A ∪B is a linear basis for V .Use this fact, in conjunction with Exercise ??, to prove the equality

(2) dim X + dim V/X = dim V.

42 Let W1 and W2 be linear subspaces of the vector space V . Let B ⊂ W1 bean arbitrary subset. Prove that the following are equivalent:

(i) B is a linear (W1 ∩W2)-basis for W1.(i) B is a linear W2-basis for W1 + W2.

Use this fact to prove the following “Paralellogram Law”

(3) dim W1 + dim W2 = dim (W1 + W2) + dim (W1 ∩W2).

Hint: Assume B satisfies condition (i). Prove first that Span(B;W2) = W1 +W2. It suffices toprove that W1 ⊂ Span(B;W2), which is pretty easy. Second, prove that B is linearly independentrelative to W2. Argue that, if v = β1b1 + · · · + βnbn ∈ W2, then in fact we have v ∈ W1 ∩W2,

which using (i) forces β1 = · · · = βn = 0. Use Exercise ?? to conclude that B is a linear W2-basisfor W1 +W2.

If B satisfies condition (ii), then it is clear that B is also linearly independent relative toW1 ∩W2. To prove the equality Span(B;W1 ∩W2) = W1, it suffices to prove only inclusion “⊃.”Start with some element w1 ∈ W1. Using (ii) there exist b ∈ Span(B) and w2 ∈ W2, such thatw1 = b+w2. This forces w2 = w1− b to belong to W1 ∩W2, so w1 ⊂ Span

(B ∪ (W1 ∩W2)

), thus

proving the desired inclusion.To prove the equality (3), we use the above equivalence, combined with exercise ?? to get

dim(W1/(W1 ∩W2)

)= Card(B) = dim

((W1 +W2)/W2

).

By adding dim(W1 ∩W2) + dimW2 to both sides, and using exercise ??, the result follows.

2. Linear maps

Definition. Suppose V and W are k-vector spaces. A map T : V → W issaid to be k-linear, if:

• T is a group homomorphism, i.e. T (x+ y) = T (x)+T (y), for all x, y ∈ V ;• T is compatible with the scalar multiplication, i.e. T (λx) = λT (x), for all

λ ∈ k, x ∈ V .

43 Suppose V is a k-vector space. For a scalar µ ∈ k, we define the multiplicationmap

Mµ : V 3 v 7−→ µv ∈ V.

Prove that Mµ is k-linear. If we take µ = 1, then Mµ = IdV , the identity map onV .

44 Let V and W be vector spaces, and let T : V →W be a linear map. For anysubset M ⊂ V , prove the equality

Span(T (M)

)= T

(Span(M)

).

45 Prove that the composition of two linear maps is again linear. Prove that theinverse of a bijective linear map is again linear. A bijective linear map is called alinear isomorphism.

2. LINEAR MAPS 11

46 Let X be a linear subspace of the vector space V , and let T : V → W be alinear map. Prove that the restriction T

∣∣X

: X →W is again a linear map.

47 Let X be a k-linear subspace of a k-vector space V . Prove that the quotientmap (defined on page 8) V 3 v 7−→ [v]X ∈ V/X is k-linear. Also prove that theinclusion ι : X → V is linear.

48 Let T : V1 → V2 be a linear map, let X1 be a linear subspace of V1, and letX2 be a linear subspace of V2. Prove the

Lemma 1.2.1. (Factorization Lemma) The following are equivalent:(i) There exists a linear map S : V1/X1 → V2/X2 such that the diagram

V1quotient map−−−−−−−−→ V1/X1

T

y yS

V2quotient map−−−−−−−−→ V2/X2

is commutative. (This means that, if we denote by qk : Vk → Vk/Xk,k = 1, 2 the quotient maps, then we have the equality q2 T = S q1.)

(ii) T (X1) ⊂ X2.Moreover, in this case the map S, described implicitly in (i), is unique.

49 Let (Vi)i∈I be a family of vector spaces. For each j ∈ I, denote by εj :Vj →

⊕i∈I Vi the standard inclusion, and by πj :

∏i∈I Vi → Vj the coordinate

projection. Prove that εj and πj are linear maps.


Theorem 1.2.1. (Universal Property of the direct sum) Let (Vi)i∈I be a familyof vector spaces. For each j ∈ I, denote by εj : Vj →

⊕i∈I Vi the standard

inclusion. Let W be a vector space, and let (Tj : Vj → W )j∈I be a collection oflinear maps. Then there exists a unique linear map T :

⊕i∈I Vi →W , such that

T εj = Tj, for all j ∈ J.

Hint: We know that any element v = (vi)i∈I ∈⊕

i∈I Vi is represented as

v =∑

j∈bvc(εj πj)(v) =

∑j∈bvc

εj(vj).

Define T (v) =∑

j∈bvc(Tj πj)(v) =∑

j∈bvc Tj(vj).

Corollary 1.2.1. Let V be a k-vector space and let W = (Wi)j∈J be a familyof linear subspaces of V . Then there exists a unique linear map ΓW :

⊕i∈I Wi → V ,

such that, for every i ∈ I, we have

(ΓW εi)(w) = w, for all w ∈Wi.

For any w = (wi)i∈I ∈⊕

i∈I Wi, we have

ΓW(w) =∑

i∈bwc

wi.


Comment. A particular case of the above result is when all the Wj ’s havedimension one (or zero). In this case the family W is represented by an elementb = (bj)j∈J ∈

∏j∈J V , by Wj = kbj . Following the above construction, we have

a linear map denoted ΓVb :

⊕j∈J k → V , which is defined as follows. For every

λ = (λj)j∈J ∈⊕

j∈J k, we have ΓVb (λ) =

∑j∈bλc λjbj ∈ V .

51 Let V be a k-vector space and letW = (Wi)j∈J be a family of linear subspacesof V . Let ΓW :

⊕j∈J Wj → V be the linear map defined in exercise ??.

A. Prove that ΓW is injective, if and only if W is a linearly independentfamily.

B. Prove that ΓW is surjective, if and only if V = Span( ⋃

j∈J Wj

).

Conclude that ΓW is an isomorphism, if and only if both condition below are true:(i) W is a linearly independent family;(ii) V = Span

( ⋃j∈J Wj

).

Definition. A family W = (Wj)j∈J of linear subspaces of V , satisfying theconditions (i) and (ii) above, is called a direct sum decomposition of V .

Comment. A particular case of the above result can be derived, along thesame lines as in the comment following exercise ??. Specifically, if V is a vectorspace, and if we have a system b = (bj)j∈J ∈

∏j∈J V , then

A. ΓVb :

⊕j∈J k→ V is surjective, if and only if V = Span

(bj : j ∈ J

).

B. ΓVb :

⊕j∈J k → V is injective, if and only if all the bj ’s are different and

the set bj : j ∈ J is linearly independent.In particular, ΓV

b :⊕

j∈J k → V is a linear isomorphism, if and only if all the bj ’sare different and the set bj : j ∈ J is a linear basis for V .

52 Let (Vj)j∈J be a direct sum decomposition for a vector space V . Let Wbe a vector space, and assume that, for each j ∈ J , we are given a linear mapTj : Vj ∈W . Prove that there exists a unique linear map T : V →W , such that

T∣∣Vj

= Tj , for all j ∈ J.

53 Let V and W be vector spaces, and let T : V →W be a linear map. Since T isa group homomorphism, we know that Ker T ⊂ V and RanT ⊂W are subgroups.Prove that they are in fact linear subspaces.

Comment. We know from group theory that the triviality of the kernel isequivalent to the injectivity of the homomorphism. As a particular case, we get thefollowing:

• A linear map is injective, if and only if its kernel is the zero subspace.

54 Generalize the above fact as follows. Let V and W be vector spaces, letT : V → W be a linear map, and let Z be a linear subspace of V . Prove that thefollowing are equivalent:

(i) The restriction T∣∣Z

: Z →W is injective.(ii) Z ∩KerT = 0.

2. LINEAR MAPS 13

55 Let X be a linear subspace of the vector space V , and let q : V → V/X denotethe quotient map.

A. Prove that Ker q = X. Use this to show that, given a linear subspace W ofV , the restriction q

∣∣W

: W → V/X is injective, if and only if W ∩X = 0.B. Given a linear subspace W of V , the restriction q

∣∣W

: W → V/X issurjective, if and only if W + X = V .

56* Prove the following technical result.

Lemma 1.2.2. (Descent Lemma) Let V be a vector space, and let T : V → Vbe a linear map. For every Define the subspaces W0 = 0 and

Wk = Ker(T · · · T︸︷︷︸k factors

),

for all k ≥ 1.(i) One has the inclusions W0 ⊂W1 ⊂W2 ⊂ . . . .(ii) For every k ≥ 1, one has Wk = T−1(Wk−1)(iii) For every k ≥ 1, there exists a unique linear map Sk : Wk+1/Wk →

Wk/Wk−1 such that the diagram

Wk+1quotient map−−−−−−−−→ Wk+1/Wk

T

y ySk

Wk −−−−−−−−→quotient map

Wk/Wk−1

is commutative.(iv) All linear maps Sk : Wk+1/Wk →Wk/Wk−1 are injective.(v) Suppose there exists some `, such that W` = W`+1. Then Wk = W`, for

all k > `.

Sketch of proof: Parts (i) and (ii) are obvious.

From (ii) we get T (Wk) ⊂ Wk−1, and then part (iii) follows from the Factorization Lemma??.

To prove part (iv), start with some x ∈ KerSk. This means that x = [w]Wk, for some

w ∈ Wk+1, and T (w) ∈ Wk−1. This forces w ∈ Wk, so x = 0 in the quotient space Wk+1/Wk.

This means that KerSk = 0, and we are done.

To prove part (v), observe first that the given condition forces W`+1/W` = 0. Use then

induction, based on (iv).

57 Let V and Z be vector spaces, let W be a linear subspace of V , and letT : W → Z be a linear map. Prove that there exists a linear map S : V → Z, suchthat S

∣∣W

= T .Hint: It suffices to prove this in the particular case when Z = W and T = Id. By exercise

??,we can choose a linear subspace X of V , with X ∩W = 0 and X + W = V . If we take

q : V → V/X the quotient map, then its restriction q∣∣W

: W → V/X is a linear isomorphism.

Put S = (q∣∣W

)−1 q.

58 Let V be a non-zero k-vector space, and let v ∈ V r 0. Prove the existenceof a k-linear map φ : V → k with φ(v) = 1.


Hint: Take W = Span(v) = kv, and define the map ψ : k 3 λ 7−→ λv ∈ W . Prove that

ψ : k → W is a linear isomorphism. Take φ0 : W → k to be the inverse of ψ, and apply the

previous exercise.

59 Let V and W be vector spaces, and let T : V → W be a linear map. For asubset M ⊂ V , prove the equality

T−1(Span

(T (M)

))= Span(M ; KerT ).

60 Let V and W be vector spaces, and let T : V → W be a linear map. For asubset M ⊂ V , prove that the following are equivalent:

(i) M is linearly independent relative to KerT .(ii) The restriction T

∣∣M

: M → W is injective and the subset T (M) ⊂ W islinearly independent.

61 Let V and W be vector spaces, and let T : V → W be a linear map. Provethat the following are equivalent:

(i) T is injective.(ii) There exists a linear basis B for V , such that the restriction T

∣∣B

: B →W

is injective, and T (B) is linearly independent.(ii’) If B is any linear basis for V , then the restriction T

∣∣B

: B → W isinjective, and T (B) is linearly independent.


(i) T is surjective.(ii) There exists a linear basis B for V , such that the restriction Span

(T (B)

)=

W . independent.(ii’) If B is any linear basis for V , Span

(T (B)

)= W .

63 Let T : V → W be a linear isomorphism from the vector space V into thevector space W . Prove that a set B ⊂ V is linear basis for V , if and only if T (B)is a linear basis for W . Conclude that dim V = dim W .


(i) T is a linear isomorphism.(ii) There exists a linear basis B for V , such that the restriction T

∣∣B

: B →Wis injective, and T (B) is a linear basis for W .

(ii’) If B is any linear basis for V , then the restriction T∣∣B

: B → W isinjective, and T (B) is a linear basis for W .

65 Let V and W be vector spaces, and let T : V → W be a linear map. Provethe following.

2. LINEAR MAPS 15

Theorem 1.2.2. (Isomorphism Theorem) There exists a unique bijective linearmap T : V/KerT → RanT such that the following diagram is commutative

Vquotient map−−−−−−−−→ V/KerT

T

y yT

W ←−−−−−inclusion

RanT

(this means that T = ι T q, where ι is the inclusion, and q is the quotient map).

Use exercise ?? to conclude that

(4) dim(Ker T ) + dim(RanT ) = dim V.

Hint: To get the existence, and uniqueness of T , use the Factorization Lemma ??, with the

subspaces KerT ⊂ V and 0 ⊂W .

66 Let V and W be finite dimensional vector spaces, with dim V = dim W , andlet T : V →W be a linear map.

A. Prove the “Fredholm alternative”

dim(KerT ) = dim (W/RanT ).

B. Prove that the following are equivalent:(i) T is a linear isomorphism.(ii) T is surjective.(iii) T is injective.

Hint: Part A follows immediately from (4) and (2). To prove the equivalence in part B, use the

fact that for a linear subspace Y ⊂W , the condition dim(W/Y ) = 0 is equivalent to the fact that

Y = W .

67* Give an infinite dimensional counterexample to the “Fredholm alternative.”Specifically, consider the infinite dimensional vetor space V =

⊕∞n=1 k and con-

struct a non-invertible surjective linear map T : V → V .

Notation. Suppose V and W are k-vector spaces. We denote by Link(V,W )the set of all k-linear maps V →W .

68 Let V and W be k-vector spaces. For T, S ∈ Lin(V,W ) we define the mapT + S : V →W by

(T + S)(v) = T (v) + S(v), for all v ∈ V.

Prove that T + S is again a linear map. Prove that Lin(V,W ) is an abelian group,when equipped with the addition operation. The neutral element is the null map0(v) = 0.

69 Let V and W be k-vector spaces. For T ∈ Lin(V,W ) and λ ∈ k, we definethe map λT : V →W by

(λT )(v) = λT (v), for all v ∈ V.

Prove that λT is a linear map. Prove that Lin(V,W ), when equipped with thismultiplication, and the addition defined above, is k-vector space.


70 The above construction is a special case of a more general one. Start with avector space W and a set X. Consider the set Map(X, W ) of all functions X →W .For f, g ∈Map(X, W ) we define f + g ∈Map(X, W ) by

(f + g)(x) = f(x) + g(x), for all x ∈ X,

and for f ∈Map(X, W ) and λ ∈ k, we define λf ∈Map(X, W ) by

(λf)(x) = λf(x), for all x ∈ X.

The Map(X, W ) becomes a k-vector space. Prove that if V is a vector space, thenLin(V,W ) is a linear subspace in Map(V,W ).

Comment. The set Map(X, W ) is precisely the product∏

x∈X W . The abovevector space structure on Map(X, W ) is precisely the direct product vector spacestructure.

71 Suppose V , W , X and Y are vector spaces, and T : X → V and S : W → Yare linear maps. Prove that the map

Lin(V,W ) 3 R 7−→ S R T ∈ Lin(X, Y )

is linear.

72 Let V be a k-vector space. Prove that the map

Link(k, V ) 3 T 7−→ T (1) ∈ V

is a k-linear isomorphism.

73 Let V and W be vector spaces. Suppose Z ( V is a proper linear subspace.Assume we have elements v ∈ V r Z and w ∈ W . Prove that there exists a linearmap T ∈ Lin(V,W ), such that T

∣∣Z

= 0, and T (v) = w.Hint: By assumption, the element v = [v]Z is non-zero in the quotient space V/Z. Use exercise

?? to find a linear map φ : V/Z → k, with φ(v) = 1. Take σ : K → W to be unique linear map

with σ(1) = w. Now we have a composition σ φ ∈ Lin(V/Z,W ), with (σ φ)(v) = w. Finally,

compose this map with the quotien map V → V/Z.

74 Suppose V is a vector space, and X is a finite set. Prove that

dim Map(X, V ) = Card(X) · dim V.

Hint: Use exercise ??, plus the identification Map(X,V ) =⊕

x∈X V .

75 Let V and W be vector spaces. Fix a subset X ⊂ V , and consider the map

ΣX : Lin(V,W ) 3 T 7−→ T∣∣X∈Map(X, W ).

Prove that ΣX is a linear map.

76* Use same notations as in the previous exercise. Assume both V and W arenon-zero.

A. Prove that ΣX is injective, if and only if V = Span(X).B. Prove that ΣX is surjective, if and only if X is linearly independent.

2. LINEAR MAPS 17

Conclude that ΣX is an isomorphism, if and only if X is a linear basis for V .Hint: A. If Span(X) ( V , choose an element v ∈ V r Span(X), and an element w ∈ W r 0.Use exercise ?? to produce a linear map T : V → W , such that T (v) = w and T

∣∣X

= 0. Such an

element is non-zero, but belongs to Ker ΣX , so ΣX is not injective. Conversely, any T ∈ Ker ΣX

will have the property that X ⊂ KerT . In particular, if Span(X) = V , this will force KerT = V ,hence T = 0.

B. Assume X is not linearly independent. This means that there exist different elementsx1, . . . , xn ∈ X and scalars α1, . . . , αn ∈ k r 0, such that α1x1 + · · ·+ αnxn = 0. Choose a anelement w ∈W r 0 and define the map f : X →W by

f(x) =

w if x = x1

0 if x 6= x1

Prove that f does not belong to Ran ΣX , so ΣX is not surjective. Conversely, assume X is linearly

independent, and let f : X → W , thought as an element b =(f(x)

)x∈X

∈∏

x∈X W . Define the

linear map ΓWb :

⊕x∈X k → W , as in exercise ??. Likewise, if we put Z = Span(X), we can use

the element a = (x)x∈X ∈∏

x∈X Z, to define a linear map ΓZa :

⊕x∈X k → Z. This time, ΓZ

a

is a linear isomorphism. If we consider the composition T = ΓWb (ΓZ

a )−1 ∈ Lin(Z,W ), we have

managed to produce a linear map with T (x) = f(x), for all x ∈ X. We now extend T to a linear

map S : V →W , which will satisfy ΣX(S) = f .

77 Suppose V and W are vector spaces.A. Prove that the following are equivalent:

(i) dim V ≤ dim W ;(ii) there exists an injective linear map T : V →W ;(iii) there exists a surjective linear map S : W → V .

B. Prove that the following are equivalent:(i) dim V = dim W ;(ii) there exists a linear isomorphism T : V →W .

Hint: Fix A a linear basis for V , and B a linear basis for W , so that ΣA : Lin(V,W ) →Map(A,W ) and ΣB : Lin(W,V ) → Map(B, V ) are linear isomorphisms. To prove A (i) ⇒ (ii),we use the definition of cardinal inequality

Card(A) ≤ Card(B) ⇔ ∃ f : A→ B injective .

For f as above, if we take T ∈ Lin(V,W ) with the property that ΣA(T ) = f , it follows that T is

injective. To prove the implication (ii) ⇒ (iii), start with some injective linear map T : V →W ,

and consider S : RanT → V to be the inverse of the isomorphism T : V → RanT . Use exercise ??to extend S to a linear map R : W → V . Since S is surjective, so is R. The implication (iii) ⇒ (i)

follows from the (2).

B. The implication (i) ⇒ (ii) has already been discussed. To prove the converse, use theequivalence

Card(A) ≤ Card(B) ⇔ ∃ f : A→ B bijective .

For f as above, if we take T ∈ Lin(V,W ) with the property that ΣA(T ) = f , it follows that T is

an isomorphism.

78 Assume V and W are vector spaces, with V finite dimensional. Prove that

(5) dim Lin(V,W ) = dim V · dim W.

As a particular case, we have

(6) dim Link(V,k) = dim V

Definition. For a k-vector space V , the vector space Link(V,k) is called thedual of V .


79 Let V be a k-vector space. Suppose B is a linear basis for V . For every b ∈ B,define the map fb : B → k by

fb =

1 if x = b0 if x 6= b

Use exercise ?? to find, for each b ∈ B, a unique element b∗ ∈ Link(V,k), withΣB(b∗) = fb. The linear map b∗ is uniquely characterized by b∗(b) = 1, andb∗(x) = 0, for all x ∈ B r b.

(i) Prove that set B∗ = b∗ : b ∈ B is linearly independent in Link(V,k).(ii) Prove that the map B 3 b 7−→ b∗ ∈ B∗ is injective, so we have Card(B) =

Card(B∗).

80 Let V be a k-vector space, and let B be a linear basis for V . Using thenotations from the preceding exercise, prove that the following are equivalent

(i) B∗ is a linear basis for Link(V,k).(ii) V is finite dimensional.

Hint: Use the isomorphism ΣB : Link(V,k) →∏

b∈B k, we have

Span(ΣB(B∗)

)= Span(fb : b ∈ B).

But when we identify Map(B,k) with∏

b∈B k, we get

Span(fb : b ∈ B) =⊕b∈B

k.

Therefore the condition Span(B∗) = Link(V,k) is equivalent to the condition∏

b∈B k =⊕

b∈B k,

which is equivalent to the condition that B is finite.

Definition. Suppose V is a finite dimensional k-vector space, and B is a linearbasis for V . Then (see the above result) the linear basis B∗ for Link(V,k) is calledthe dual basis to B. In fact the notion of the dual basis includes also the bijectionB 3 b 7−→ b∗ ∈ B∗ as part of the data.

3. The minimal polynomial

Notations. From now on, the composition of linear maps will be writtenwithout the symbol. In particular, if V is a k-vector space, if n ≥ 2 is andinteger, and if T ∈ Link(V, V ), we write Tn instead of T T · · · T (n factors).Instead of Link(V, V ) we shall simply write Link(V ).

81 Suppose V is a k-vector space. The composition operation in Link(V, V ) isobviously associative. Prove that, for every T ∈ Link(V ), the maps

Link(V ) 3 S 7−→ ST ∈ Link(V )Link(V ) 3 S 7−→ TS ∈ Link(V )

are linear. (In particular, the composition product is distributive with respect tothe addition.)

Comment. The above result states that Link(V ) is a unital k-algebra. Theterm “unital” refers to the existence of a multiplicative identity element. In ourcase, this is the identity IdV , which from now on will be denoted by I (or IV whenwe need to specify the space).

3. THE MINIMAL POLYNOMIAL 19

82 If dim V ≥ 2, prove that the algebra Lin(V ) is non-commutative, in the sensethat there exist elements T, S ∈ Lin(V ) with TS 6= ST . (Give an example whenTS = 0, but ST 6= 0.

83 Let V be a vector space, and let S, T : V → V be linear maps. Suppose Sand T commute, i.e. ST = TS. Prove

(i) For any integers m,n ≥ 1, the linear maps Sm and Tn commute: SmTn =TnSm.

(ii) The Newton Binomial Formula holds:

(S + T )n =n∑

k=0

(nk

)Sn−kT k.

Hints: For (i) it Suffices to analyze the case n = 1. Use induction on m

For (ii) use (i) and induction.

The following is a well-known result from algebra

Theorem 1.3.1. Suppose L is a unital k-algebra. Then for any element X ∈ L,there exists a unique unital k-algebra homomorphism ΦX : k[t] → L, such thatΦX(t) = X.

Here k[t] stands for the algebra of polynomials in a (formal) variable t, withcoefficients in k. The unital algebra homomorphism condition means that

• ΦX is linear;• ΦX is multiplicative, in the sense that φx(PQ) = ΦX(P )ΦX(Q), for all

p, q ∈ k[t];• Φx(1) = I.

To be more precise, the homomorphism ΦX is constructed as follows. Start withsome polynomial P (t) = α0 + α1t + · · ·+ αntn. Then

ΦX(P ) = α0I + α1X + · · ·+ αnXn.

Definition. Suppose V is a k-vector space, and X : V → V is a linearmap. If we apply the above Theorem for the algebra Link(V ), and the element X,the corresponding map ΦX : k[t] → Link(V ) is called the polynomial functionalcalculus of X. For a polynimial P ∈ k[t], the element Φx(P ) will simply be denotedby P (X).

Remark 1.3.1. If P and Q are polynomials, then P (X)Q(X) = Q(X)P (X),simply because both are equal to (PQ)(X). This means that, although Link(V ) isnot commutative, the sub-algebra

P (X) : P ∈ k[t] ⊂ Link(V )

is commutative.

84 Let V be a vector space, and let S, T : V → V be linear maps. Assume Sand T commute. Prove that, for any two polynomials P,Q ∈ k[t], the linear mapsP (S) and Q(T ) commute.

85 Prove that the algebra k[t] is infinite dimensional, as a k-vector space. Moreexplicitly, the set 1 ∪ tn : n ≥ 1 is a linear basis for k[t].


86 Let V be a finite dimensional k-vector space. For every X ∈ Link(V ), provethat there exists a non-zero polynomial P ∈ k[t], such that P (X) = 0.Hint: Consider the polynomial functional calculus ΦX : k[t] → Link(V ). On the one hand, we

notice that dimLink(V ) = (dimV )2 < ∞ (by exercise ??). This will force dim(Ran ΦX) < ∞.

On the other hand, we know that

dim(KerΦX) + dim(Ran ΦX) = dim(k[t]).

But since k[t] is infinite dimensional, this forces KerφX to be infinite dimensional. In particular

KerΦX 6= 0.

87 Suppose V is a finite dimensional k-vector space, and X ∈ Link(V ). Provethat there exists a unique polynomial M ∈ k[t] with the following properties:

(i) M(X) = 0.(ii) M is monic, in the sense that the leading coefficient is 1.(iii) If P ∈ k[t] is any polynomial with P (X) = 0, then P is divisible by M

Hint: One (conceptual) method of proving this result is to quote the property of k[t] of being a

principal ideal domain. Then KerΦX , being an ideal, must be presented as KerΦX = M · k[t],

for a unique monic polynomial M .Here is the sketch of a “direct” proof (which in fact traces the steps used in proving that k[t]

is a pid, in this particular case). Choose a (non-zero) polynomial N ∈ k[t], of minimal degree,such that N(X) = 0. Replace N with M = λN , so that M is monic (we will still have M(X) = 0).Suppose now P is a polynomial with P (x) = 0. Divide P by M with remainder, so that we have

P = MQ+R, for some polynomials Q and R, with degR < degM . Using the fact that M(X) = 0,we get

R(X) = M(X)Q(X) +R(X) = P (X) = 0.

By minimality, this forces R = 0, thus proving property (iii). The uniqueness is a direct application

of (iii).

Definition. Suppose V is a finite dimensional k-vector space, and X : V → Vis a linear map. The polynomial M , described in the preceding exercise, is calledthe minimal polynomial of X, and will be denoted by MX .

88* Give an infinite dimensional counter example to exercise ??.Hint: Take V = k[t] and define the linear map X : k[t] 3 P (t) 7−→ tP (t) ∈ k[t]. Prove that

there is no polynomial M such that M(X) = 0.

89 Let V be a finite dimensional vector space, and fix some X ∈ Lin(V ). Provethat the following are equivalent:

• X is a scalar multiple of the identity, i.e. there exists some α ∈ k suchthat X = αI.• The minimal polynomial MX(t) is of degree one.

Definition. Suppose V is a vector space. A linear map X : V → V is said tobe nilpotent, if there exists some integer n ≥ 1, such that Xn = 0.

90 Let V be a finite dimensional vector space, and let X ∈ Lin(V ) be nilpotent.Prove that the minimal polynomial is of the form MX(t) = tp, for some p ≤ dim V .In particular, we have Xdim V = 0.Hint: Fix n such that Xn = 0 Consider the linear subspaces W0 = 0 and Wp = Ker(Xk),

k ≥ 1. It is clear that MX(t) = tp, where p is the smallest index for which Wp = V . Prove that


for every k ∈ 1, . . . , p we have the strict inclusion Wk−1 ( Wk. (Use the Descent Lemma ??)Use then (recursively) the dimension formula for quotient spaces (2), to get

dimV =

p∑k=1

dim(Wk/Wk−1).

Since each term in the sum is ≥ 1, this will force dimV ≥ p.

91 Let V be a vector space, and let M,N : V → V be nilpotent linear maps. IfM and N commute, then prove that M +N is again nilpotent. If X : V → V is anarbitrary linear map, which commutes with N , then prove that XN is nilpotent.

92* Suppose V is finite dimensional, X ∈ Lin(V ), and W is an invariant linearsubspace, i.e. with the property T (W ) ⊂ W . Let T : V/X → V/X be the uniquelinear map which makes the digram

Vquotient map−−−−−−−−→ V/W

X

y yT

Vquotient map−−−−−−−−→ V/W

commutative (see exercise ??). Let us also consider the linear map S = X∣∣W

:W →W . Prove that

(i) The minimal polynomial MX(t) of X divides the product of minimal poly-nomials MS(t)MT (t).

(ii) The polynomial MX(t) is divisble by both minimal polynomials MS(t) andMT (t), hence MX(t) is divisible by the lowest common multiple (lcm) ofMS(t) and MT (t).

Conclude that, if MS(t) and MT (t) are relatively prime, then

MX(t) = MS(t)MT (t).

Hint: Observe first that, for any polynomial P ∈ k[t], one has a commutative diagram

Vquotient map−−−−−−−−−→ V/W

P (X)

y yP (T )

Vquotient map−−−−−−−−−→ V/W

and the equality P (S) = P (X)∣∣W

. In particular, since MX(X) = 0, we immediately get MX(S) =

0 and MX(T ) = 0, thus proving (ii). To prove (i), we only need to show that MS(X)MT (X) = 0.

Start with an arbitrary element v ∈ V . Since MT (T ) = 0, using the above diagram we get theequality

q([MT (X)](v)

)= 0, in V/W,

which means that the element w = [MT (X)](v) belongs to W . But then, using MS(S) = 0, weget

0 = [MS(S)](w) = [MS(X)](w) =([MS(X)]

([MT (X)](v)

))= [MS(X)MT (X)](v).

93 The following example shows that, in general the minimal polynomial MX(t)can differ from both the product MS(t)MT (t) and lcm(MS(t),MT (t)). Considerthe space V = k4 and the linear map

X : k4 3 (α1, α2, α3, α4) 7−→ (0, α1, 0, α2) ∈ k4.


Take the linear subspace W = (0, 0, λ, µ) : λ, µ ∈ k ⊂ V . Check the equalities:MS(t) = t, MT (t) = t2, and MX(t) = t4.Hint: Prove that S = 0, T 2 = 0, but T 6= 0, and X4 = 0, but X3 6= 0.

94* Suppose V is finite dimensional, X ∈ Lin(V ), and W1, . . . ,Wn are invariantsubspaces, such that W1 + · · ·+ Wn = V . For each k = 1, . . . , n we take Mk(t) tobe the minimal polynomial of X

∣∣Wk. Prove that the minimal polynomial of X isgiven as

MX(t) = lcm(M1(t), . . . ,Mk(t)).

Hint: Use induction on k. In fact, it suffices to prove the case k = 2. As in the previ-

ous exercise, prove that MX(t) is divisible by lcm(M1(t), . . . ,Mk(t)). Conversely, if we de-note lcm(M1(t), . . . ,Mk(t)) simply by P (t), we know that for every k, we have a factorization

P (t) = Qk(t)Mk(t). In particular, for every w ∈Wk, we get

[P (X)](w) = [Qk(X)]([Mk(X)](w)

)= [Qk(X)]

([Mk(X

∣∣Wk

)](w))

= 0.

Now we have P (X)∣∣Wk

= 0, for all k, which gives P (X) = 0, thus proving that P (t) is divisible

by MX(t).

95* Let V be a finite dimensional k-vector space, and let X : V → V be a linearmap. Prove the following

Theorem 1.3.2. (Abstract Spectral Decomposition) Assume the minimal poly-nomial is decomposed as

(7) MX(t) = M1(t) ·M2(t) · · · · ·Mn(t),

such that for any j, k ∈ 1, . . . , n with j 6= k, we have gcd(Mj ,Mk) = 1. Definethe polynomials

Pj(t) =MX(t)Mj(t)

, j = 1, . . . , n.

Since gcd(P1, P2, . . . , Pn) = 1, we know there exist polynomials Q1, . . . , Qn ∈ k[t],such that

(8) Q1(t)P1(t) + Q2(t)P2(t) + · · ·+ Qn(t)Pn(t) = 1.

Define the linear maps

Ej = Qj(X)Pj(X) ∈ Link(V ), j = 1, 2, . . . , n.

(i) The linear maps Ej, j = 1, . . . , n are idempotents, i.e. Ej2 = Ej.

(ii) For any j, k ∈ 1, . . . , n with j 6= k, we have EjEk = 0.(iii) E1 + E2 + · · ·+ En = I.(iv) For each j ∈ 1, . . . , n we have the equality RanEj = Ker[Mj(X)].(v) For every j, the minimal polynomial of the restriction X

∣∣Ran Ej

is preciselyMj(t).

Sketch of proof: By construction, we have (iii), so in fact conditions (i) and (ii) are equivalent.To prove (ii) start with j 6= k. Using the fact that PjPk is divisible by all the Mi’s it follows that

PjPk is divisible by MX , so we get Pj(X)Pk(X) = 0. Using polynomial functional calculus, we

get

EjEk = Qj(X)Pj(X)Qk(X)Pk(X) = Qj(X)Qk(X)Pj(X)Pk(X) = 0.

To prove (iv) we first observe that since Mj(t)Pj(t) = MX(t), we have (again by functional

calculus)

Mj(X)Ej = Mj(X)Qj(X)Pj(X) = Qj(X)Mj(X)Pj(X) = Qj(X)MX(X) = 0,


which proves the inclusion RanEj ⊂ Ker[Mj(X)]. Conversely, since Pk(t) is divisible by Mj(t)

for all k 6= j, we see that Pk(X)∣∣Ker[Mj(X)]

= 0, which forces Ek

∣∣Ker[Mj(X)]

= 0, for all k 6= j.

Using (iii) this gives Ej

∣∣Ker[Mj(X)]

= Id∣∣Ker[Mj(X)]

, thus proving the other inclusion.

To prove (v), denote for simplicity the subspaces RanEj by Wj . By (iii) we know that

W1 + · · ·+Wn = V . By exercise ?? we know that

(9) MX = lcm(M01 , . . . ,M

0n),

where M0j is the minimal polynomial of X

∣∣Wj

. By (iii) we know that Mj(X∣∣Wj

) = 0, hence Mj(t)

is divisble by M0j (t). In particular we will also have gcd(M0

j ,M0k ) = 1, for all j 6= k, so by (9) we

will get

MX(t) = M01 (t)M0

2 (t) · · · · ·M0n(t).

This forces M0j = Mj , for all j

Definition. Use the hypothesis of the above Theorem. The idempotentsE1, . . . , En are called the spectral idempotents of X, associated with the decom-position (7). The subspaces RanEj = Ker[Mj(X)], j = 1, . . . , n are called thespectral subspaces of X, associated with the decomposition (7). Notice that

(i) V = Ker[M1(X)] + Ker[M2(X)] + · · ·+ Ker[Mn(X)].(ii)

(Ker[Mj(X)]

)n

j=1is a linearly independent family of linear subspaces.

In other words,(Ker[Mj(X)]

)n

j=1is a direct sum decomposition of V .

96 Prove the properties (i) and (ii) above.

97 The above definition to carries a slight ambiguity, in the sense that thespectral idempotents Ej are defined using a particular choice of the polynomialsQj for which (8) holds. Prove that if we choose another sequence of polynomials,Q1, . . . , Qn, such that

Q1(t)P1(t) + Q2(t)P2(t) + · · ·+ Qn(t)Pn(t) = 1,

then we have the equalities

Qj(X)Pj(X) = Ej , for all j = 1, . . . , n.

Therefore the spectral idempotents are un-ambiguously defined.Hint: Define Ej = Qj(X)Pj(X), j = 1, . . . , n. Use the Theorem to show RanEj = RanEj , and

then use the Theorem again to show that Ej = Ej , for all j.

Another approach is to look at the spectral decomposition

V = Ker[M1(X)] + Ker[M2(X)] + · · ·+ Ker[Mn(X)],

which depends only on the factorization (7). For every element v ∈ V , there are unique elements

wj ∈ Ker[Mj(X)], j = 1, . . . , n, such that v = w1 + · · ·+ wn. Then Ej(v) = wj .

98 Prove the following

Theorem 1.3.3. (Uniqueness of Spectral Decomposition) Let V be a finitedimensional vector space, let X : V → V be a linear map, and let W1, . . . ,Wn be acollection of invariant linear subspaces. For each j = 1, . . . , n, define Mj(t) to bethe minimal polynomial of X

∣∣Wj∈ Lin(Wj). Assume that:

(i) W1 + · · ·+ Wn = V .(ii) For every j, k ∈ 1, . . . , n with j 6= k, we have gcd(Mj ,Mk) = 1.


Then the minimal polynomial decomposes as MX(t) = M1(t) · · · · · Mn(t), andmoreover, the Wj’s are precisely the spectral subspaces associated with this decom-position, i.e. Wj = Ker[Mj(X)], for all j = 1, . . . , n.

Hint: The decomposition of MX(t) has already been discussed (see exercise ??). It is clear

that Wj ⊂ Ker[Mj(X)]. On the one hand, (Wj)nj=1 is a linearly independent family of linear

subspaces, so

dimV = dim(W1 + · · ·+Wn) = dimW1 + · · ·+ dimWn.

On the other hand, we have dimWj ≤ dim(Ker[Mj(X)]), and we also have

dimV = dim(Ker[M1(X)]) + · · ·+ dim(Ker[Mn(X)]).

By finiteness, this forces dimWj = dim(Ker[Mj(X)]), so we must have Wj = Ker[Mj(X)].

99* Work again under the hypothesis of the Abstract Spectral DecompositionTheorem. The following is an alternative construction of the spectral idempotents.Suppose R1, . . . , Rn are polynomials such that, for each j ∈ 1, . . . , n, we have:

(i) Rj is divisible by Mk, for all k ∈ 1, . . . , n, k 6= j;(ii) 1−Rj is divisible by Mj .

(See the hint on why such polynomials exist.) Prove that Rj(X) = Ej , for all j.Hint: The simplest example of the existence of R1, . . . , Rn is produced by taking Rj(t) =Qj(t)Pj(t), j = 1, . . . , n.

Define Fj = Rj(X). On the one hand, j, k ∈ 1, . . . ,m with j 6= k, the polynomial RjRk is

divisible by both Pj and Pk, hence it is divisible by gcd(Pj , Pk) = MX . In particular we get

FjFk = Rj(X)Rk(X) = (RjRk)(X) = 0.

On the other hand, for a fixed j, it is clear that Rk is divisible by Mj , for all k 6= j. Therefore,

if we consider the polynomial R(t) = R1(t) + · · ·+Rn(t), we see that R−Rj is divisible by Mj .

But so is 1− Rj . This means that 1− R is divisble by Mj . Since this is true for all j, it followsthat 1 − R is divisible by gcd(M1, . . . ,Mn) = MX . This forces R(X) = 1, which means that

F1 + · · ·+ Fn = I, so we must have Fj2 = Fj .

Now we have a linearly independent family of linear subspaces(RanFj

)n

j=1, with

(10) RanFn + · · ·+ RanFn = V.

Notice that MjRj is divisible by MX , so Mj(X)Fj = Mj(X)Rj(X) = 0. This means that

RanFj ⊂ Ker[Mj(X)]. Because of (10), this forces RanFj = Ker[Mj(X)] = RanEj , which in

turn forces Fj = Ej .

100 Suppose V and W are k-vector spaces, and S : V → W is a linear isomor-phism. Prove that the map

Link(V ) 3 X 7−→ SXS−1 ∈ Link(W )

is an isomorphism of k-algebras. Use this fact tow prove that, for a linear mapX ∈ Link(V ) one has

P (SXS−1) = SP (X)S−1, for all P ∈ k[t].

101 Suppose V , W and S are as above, and V (hence also W ) finite dimensional.Prove that for any X ∈ Lin(V ) we have the equality of minimal polynomialsMX(t) = MSXS−1(t).

4. SPECTRUM 25

4. Spectrum

Convention. Throughout this section we restrict ourselves to a particularchoice of the field k = C, the field of complex numbers. All vector spaces areassumed to be over C.

This choice of the field is justified by the fact that the arithmetic in C[t] isparticularly nice, as shown by the following:

Theorem 1.4.1. (Fundamental Theorem of Algebra) For every non-zero poly-nomial P ∈ C[t] there exists a collection of different numbers λ1, . . . , λn ∈ C, andintegers m1, . . . ,mk ≥ 1, such that

P (t) = (t− λ1)m1 · · · · · (t− λn)mn .

The numbers λ1, . . . , λn are called the roots of P . They are precisely thesoultions (in C) of the equation P (λ) = 0. The numbers m1, . . . ,mn are called themultiplicities. More precisely, mk is called the multiplicity of the root λk. Remarkthat

deg P = m1 + · · ·+ mn.

Definitions. Let V be a vector space, and X : V → V be a linear map. Thespectrum of X is defined to be the set

Spec(X) = λ ∈ C : MX(λ) = 0.

Suppose λ ∈ Spec(X). The multiplicity of λ, as a root of the minimal polynomial, iscalled the spectral multiplicity of λ, and is denoted by mX(λ). The linear subspace

N(X, λ) = Ker[(X − λI)mX(λ),

is called the spectral subspace of λ. The number

dX(λ) = dim N(X, λ)

is called the spectral dimension of λ.

According to the Fundamental Theorem of Algebra, if we list the spectrum asSpec(X) = λ1, . . . , λn, with the λ’s different, we have a decomposition

(11) MX(t) = (t− λ1)mX(λ1) · · · · · (t− λn)mX(λn).

When we apply the Abstract Spectral Decomposition Theorem for the polynomialsMj(t) = (t− λj)mX(λj), j = 1, . . . , n, we get the following

Theorem 1.4.2. (Spectral Decomposition Theorem) Suppose V is a finitedimensional vector space, and X : V → V is a linear map. List the spectrum asSpec(X) = λ1, . . . , λn, with all λ’s different.

(i) The spectral subspaces N(X, λj), j = 1, . . . , n, form a linearly independentfamily of linear subspaces.

(ii) N(X, λ1) + · · ·+ N(X, λn) = V .

If for every j ∈ 1, . . . , n we define Xj = X∣∣N(X,λj)

∈ Lin(N(X, λj)

), then

(iii) the minimal polynomial of Xj is (t−λj)mX(λj); in particular the spectrumis a singleton Spec(Xj) = λj.


Definitions. Suppose V is a finite dimensional vector space, and X : V → Vis a linear map. The family

(N(X, λ)

)λ∈Spec(X)

, which by Theorem ?? is a directsum decomposition of V , is called the spectral decomposition of X. The spectralidempotents of X are the ones described in the Abstract Spectral DecompositionTheorem (exercise ??), corresponding to (11). They are be denoted by EX(λ),λ ∈ Spec(X). These idempotents have the following properties (which uniquelycharacterize them):

(i) EX(λ)EX(µ) = 0, for all λ, µ ∈ Spec(X) with λ 6= µ;(ii) Ran EX(λ) = N(X, λ), for all λ ∈ Spec(X).

As a consequence, we also get∑

λ∈Spec(X) EX(λ) = I.

102 Let V be a finite dimensional vector space, and let X : V → V be a linearmap. Prove that, for every λ ∈ Spec(X), we have the inequality mX(λ) ≤ dX(λ).

Hint: Consider the linear map Xλ = X∣∣N(X,λ)

∈ Lin(N(X,λ)

). We know that the minimal

polynomial ofXλ isMλ(t) = (t−λ)mX (λ). Then the linear map T = Xλ−λI : N(X,λ) → N(X,λ)

is nilpotent. Use exercise ?? to conclude that (Xλ − λI)dX (λ) = 0, hence obtaining the fact that

(t− λ)dX (λ) is divisible by Mλ(t).

Definition. Let V be a finite dimensional vector space, let X : V → V be alinear map, and let Spec(X) = λ1, . . . , λn be a listing if its spectrum (with allλ’s different). The polynomial

HX(t) = (t− λ1)dX(λ1) · · · · · (t− λn)dX(λn)

is called the chracteristic polynomial of X.

Remark 1.4.1. The above result gives the fact that the minimal polynomialMX divides the characteristic polynomial HX .

103 Prove that deg HX = dim V . As a consequence of the above remark, wehave.

Proposition 1.4.1. The degree of the minimal polynomial does not exceeddim V .

Hint: Use the Spectral Decomposition to derive the fact that degHX = dimV .

104 Suppose V and W are finite dimensional vector spaces and S : V → Wis a linear isomorphism. Prove that for any X ∈ Lin(V ) we have the equality ofcharacteristic polynomials HX(t) = HSXS−1(t).Hint: Using the equality of minimal polynomialsMX(t) = MSXS−1 (t), we know that Spec(X) =Spec(SXS−1). Prove that, for every λ ∈ Spec(X) we have

S(N(X,λ)

)= N(SXS−1, λ),

hence dX(λ) = dSXS−1 (λ).

105 Prove the following characterization of the spectrum.

Proposition 1.4.2. Let V be a finite dimensional vector space, and X : V →V be a linear map. For a complex number λ, the following are equivalent:

(i) λ ∈ Spec(X);

4. SPECTRUM 27

(ii) the linear map X − λI is not injective;(iii) the linear map X − λI is not surjective;(iv) the linear map X − λI is not bijective;

Hint: The equivalence of (ii)-(iv) follows from the Fredholm alternative. If λ ∈ Spec(X), uponwriting MX(t) = P (t)(t − λ), we see that 0 = MX(X) = P (X)(X − λI), so X − λI cannot beinvertible. Conversely, if λ 6∈ Spec(X), then when we divide MX(t) by t − λ we get a non-zeroremainder. In other words, we get MX(t) = (t − λ)Q(t) + µ, for some polynomial Q and some

µ 6= 0. But then, using functional calculus, we get

0 = MX(X) = (X − λI)Q(X) + µI,

which proves that the linear map Y = −µ−1Q(X) satisfies (X−λ)Y = Y (X−λI) = I, so X−λIis invertible.

The above characterization, specifically condition (ii) brings up an importantconcept.

Definition. A number λ is said to be an eigenvalue for the linear map X :V → V , if there exists a non-zero element v ∈ V , with X(v) = λv. Such a v is thencalled an eigenvector for X, corresponding to λ.

With this terminiology, we have λ ∈ Spec(X), if and only if λ is an eigenvaluefor X. (This condition is equivalent to the fact that Ker(X − λI) 6= 0, which isthe same as the non-injectivity of X − λI.)


Theorem 1.4.3. (Spectral Mapping Theorem) If V is a finite dimensionalvector space, and X : V → V is linear, then for any polynomial P ∈ C[t] one hasthe equality

Spec(P (X)

)= P

(Spec(X)

).

Sketch of proof: To prove the inclusion “⊃” start with some λ ∈ Spec(X), and use exercise?? to get a λ-eigenvector v for X. Then show that v is a P (λ)-eigenvector for P (X), hence

P (λ) ∈ Spec(P (X)

).

To prove the inclusion “⊂” start with some µ ∈ Spec(P (X)

), and Use the Fundamental

Theorem of Algebra to get a factorization

P (t)− µ = (t− λ1) · (t− λn),

for some λ− 1, . . . , λn ∈ C (not necessarily different). Use functional calculus to obtain

P (X)− µI = (X − λ1I) · (X − λnI).

Argue that at least one of the linear maps X − λjI, j = 1, . . . , n is non-invertible, so that at least

one of the λj ’s belongs to Spec(X). For such a λj we then have µ = P (λj) ∈ P(Spec(X)

).

The Spectral Mapping Theorem has numerous applications. One of them isthe following.

107 Let V be a finite dimensional vector space, and let X : V → V be a linearmap. Consider the linear map S : V → V defined by

S =∑

λ∈Spec(X)

λEX(λ),

where EX(λ), λ ∈ Spec(X), are the spectral idempotents of X. Prove that


(i) The minimal polynomial of S is

MS(t) =∏

λ∈Spec(X)

(t− λ).

In particular MS(t) has simple roots.(ii) The linear map N = X − S is nilpotent.

Hints: Let us recall (see exercise ??) that the spectral idempotents can be constructed asEX(λ) = Rλ(X), where Rλ(t), λ ∈ Spec(X) is a family of polynomials with the following prop-

erties:

(α) Rλ(t) is divisible by (t− µ)mX (µ), for all λ, µ ∈ Spec(X) with λ 6= µ;

(β) 1−Rλ(t) is divisible by (t− λ)mX (λ), for all λ ∈ Spec(X).

In particular, we get

(12) Rλ(µ) = δλµ, for all λ, µ ∈ Spec(X).

Now, by functional calculus, the linear map S is given by S = P (X), where

P (t) =∑

λ∈Spec(X)

λRλ(t),

and then (12) will give

(13) P (λ) = λ, for all λ ∈ Spec(X).

Using the Spectral Mapping Theorem, we get the equality Spec(S) = Spec(X). Notice that, by

construction, the subspaces Wλ = RanEX(λ) = Ker[(X − λI)mX (λ)], λ ∈ Spec(X) = Spec(S),

for a direct sum decomposition of V . It is pretty obvious that S∣∣Wλ

= λIWλ, for all λ ∈ Spec(S),

which means that the minimal polynomial of S∣∣Wλ

is Mλ(t) = t− λ. Using exercise ?? we then

have the desired form for MS(t).If we look at N = X −S, we see that we can write, N = Q(X), where Q(t) = t−P (t). Then

using (13) we have

Q(λ) = 0, for all λ ∈ Spec(X),

so by the Spectral Mapping Theorem we get Spec(N) = 0, so N is nilpotent.

Definition. Let V be a finite dimensional vector space. A linear map S :V → V is said to be semisimple, if the minimal polynomial MS(t) has simple roots.

The above result then states that any linear X : V → V map has a “Semisimple+ Nilpotent” decomposition X = S + N , with both S and N representable usingfunctional calculus of X.

108 Let V be a finite dimensional vector space, and let X : V → V be a linearmap. Prove that the following are equivalent:

(i) X is semisimple.(ii) There exist idempotents E1, . . . , En, with

• EiEj = 0, for all i, j ∈ 1, . . . , n, with i 6= j,• E1 + · · ·+ En = I,

and there exist complex numbers λ1, . . . , λn ∈ C, such that

X = λ1E1 + · · ·+ λnEn.

(iii) There exists a linear basis for V consisting of eigenvectors for X.Hint: Suppose X is semisimple. Use the construction given in exercise ?? to produce thesemisimple map

S =∑

λ∈Spec(X)

λEX(λ).

We know that S can be written as S = P (X), where P ∈ C[t] is a certain polynomial, with

P (λ) = λ, for all λ ∈ Spec(X). In particular, the polynomial Q(t) = t − P (t) is divisible by

4. SPECTRUM 29

∏λ∈spec(X)(t−λ) = MX(t), hence Q(X) = 0, which by functional calculus gives X = P (X) = S,

thus proving (ii).Assume now X satisfies (ii). In particular the subspaces Wj = RanEj , j = 1, . . . , n form a

direct sum decomposition for V . If, for every j ∈ 1, . . . , n, we choose a basis Bj for Wj , thenB =

⋃nj=1Bλ is a desired linear basis (each b ∈ Bj is a λj-eigenvector for X).

Finally, assume B is a linear basis for V consisting of eigenvectors for S. This means thatwe have a map B 3 b 7−→ θ(b) ∈ C, such that

S(b) = θ(b)b, for all b ∈ B.

Put Λ = θ(b) : b ∈ B. For each λ ∈ Λ, we define the set Bλ = θ−1(λ), and we put

Wλ = Span(Bλ). Then it is clear that S∣∣Wλ

= λIWλ, for all λ ∈ Λ, so that the minimal

polynomial of S∣∣Wλ

is Mλ(t) = t − λ. Since B =⋃

λ∈ΛBλ, we have∑

λ∈ΛWλ = V , so by

exercise ?? we conclude that the minimal polynomial of S is MS(t) =∏

λ∈Λ(t− λ).

109 Let V be a finite dimensional vector space, and let S, T : V → V besemisimple linear maps, which commute, i.e. ST = TS. Prove that S + T and STare again semisimple.Hints: Use the spectral decompositions to write S =

∑λ∈Spec(S) λES(λ), with the ES(λ)’s

obtained from functional calculus of S. Likewise, we have the spectral decomposition T =∑µ∈Spec(T ) λET (µ), with the ET (µ)’s obtained from functional calculus of T . In particular,

the ES(λ)’s commute with the ET (µ)’s. Using this fact we see that, if we define the linear mapsFλµ = ES(λ)ET (µ), then the Fλµ’s will be idempotents. Moreover, we will have

• Fλ1µ1Fλ2µ2 = 0, if (λ1, µ1) 6= (λ2, µ2);

•∑

λ∈Spec(S)µ∈Spec(T )

Fλµ = I.

Now we have

T + S =∑


(λ+ µ)Fλµ, TS =∑


(λµ)Fλµ,

and we can apply exercise ??.

110 The result from exercise ?? can be strengthened. Prove the following.

Theorem 1.4.4. (“Semisimple + Niplotent” Theorem) Let V be a finite di-mensional vector space, and let X : V → V be a linear map. Then there exists aunique decomposition X = S + N , with

(i) S semisimple;(ii) N niplotent;(iii) S and N commute, i.e. SN = NS.

Skecth of proof: The existence is clear from exercise ??.To prove uniqueness, assume X = S′ + N ′ is another decomposition, with properties (i)-

(iii). Notice that both S′ and N ′ commute with X, hence they also commute with the spectral

idempotents EX(λ), λ ∈ Spec(X). In particular, S′ and S commutive, and also N ′ and Ncommute. Now we have

S′ − S = N −N ′,

with the lefthand side semisimple (exercise ??) and the righthand nilpotent (exercise ??). But the

only linear map, which is both semisimple and nilpotent, is the null map. This forces S′ = S and

S = N .


5. The similarity problem and the spectral invariant

Definition. Let V be a vector space, and let X, Y : V → V be linear maps.We say that X is similar to Y , if there exists a linear isomorphism S : V → V , suchthat Y = SXS−1.

111 Prove that “is similar to” is an equivalence relation.

The following is a fundamental question in linear algebra.

Similarity Problem. Suppose V is a finite dimensional vector space, andX, Y : V → V are linear maps. When is X similar to Y ?

An additional problem is to construct at least one linear isomorphism S : V →V , such that Y = SXS−1.

The approach to the Similarity Problem is by constructing various (computable)invariants. More precisely, for every linear map X : V → V , we want to constructan “object” ΘX , such that

• if X is similar to Y , then ΘX = ΘY .If we have such a construction, we call the correspondence Lin(V ) 3 X 7−→ ΘX asimilarity invariant. Then our main goal will be to construct a “strong” similarityinvariant, in the sense that

• If ΘX = ΘY , then X is similar to Y .In this section we will eventually construct such a strong invariant for complexvector spaces.

Convention. All vector spaces in this section are assumed to be over C.

Examples. The following are examples of similarity invariants(i) the spectrum Lin(V ) 3 X 7−→ Spec(X);(ii) the minimal polynomial Lin(V ) 3 X 7−→MX ∈ C[t];(iii) the characteristic polynomial Lin(V ) 3 X 7−→ HX ∈ C[t].

Note that (ii) and (iii) are stronger than (i).

112 The mininmal and the charecteristic polynomial are not strong enough.Give examples of linear maps T,X ∈ Lin(C3), and Y, Z ∈ Lin(C2) such thatMT (t) = MX(t), HY (t) = HZ(t), but HT (t) 6= HX(t) and MY (t) 6= MZ(t). Thisshows that T is not similar to X, although MT (t) = MX(t), and Y is not similarto Z, although HY (t) = HZ(t).Hint: Take T (α1, α2, α3) = (0, α2, α3), and X(α1, α2, α3) = (0, 0, α3).

Take Y (α1, α2) = (0, α1), Z(α1, α2) = (0, 0).

113 Even when the minimal and characteristic polynomial are combined, i.e.when we consider the invariant ΘX = (MX ,HX), we still do not get a stronginvariant. Consider the linear maps X, Y ∈ Lin(C4), defined by

X(α1, α2, α3, α4) = (0, α1, 0, α3)Y (α1, α2, α3, α4) = (0, α1, 0, 0)

Later on we shall see why X and Y are not similar. For the moment prove thatMX(t) = MY (t) = t2, and HX(t) = HY (t) = t4.

5. THE SIMILARITY PROBLEM AND THE SPECTRAL INVARIANT 31

Definition. Let V be a finite dimensional vector space, and let X : V → Vis a linear map. For λ ∈ Spec(X), we define the linear subspaces Nk(X, λ) =Ker[(X − λI)k], k ≥ 1. These subspaces are called the generalized eigenspaces ofX, corresponding to λ. More specifically, Nk(X, λ) will be called the generalizedeigenspace of order k. It will be convenient to use the convention N0(X, λ) = 0.

114 Suppose V is a finite dimensional vector space, and X : V → V is a linearmap. Fix λ ∈ Spec(X). Prove that for any integers p > k ≥ 1 one has the inclusion

Xk(Np(X, λ)

)⊂ Np−k(X, λ).

115 Suppose V is a finite dimensional vector space, and X : V → V is a linearmap. Fix λ ∈ Spec(X). Prove that the sequence of generalized λ-eigenspacesexhibits the following inclusion pattern:

(14) 0 ( N1(X, λ) ( · · · ( NmX(λ)(X, λ) = NmX(λ)+1(X, λ) = · · · = N(X, λ).

Hint: It is clear that we have the inclusions

0 ( N1(X,λ) ⊂ N2(X,λ) ⊂ . . . ,

Define the maps Yk = X∣∣Nk(X,λ)

∈ Lin(Nk(X,λ)

). On the one hand, it is clear that (Yk−λI)k =

0, so that if we denote by Mk the minimal polynomial of Yk, we know that

(α) The polynomial Mk(t) divides the polynomial (t− λ)k.

On the other hand, by exercise ?? we also know that the minimal polynomial Mk of Yk alsodivides the minimal polynomial MX of X. Combined with (α) this gives

(β) The polynomial Mk(t) divides the polynomial (t− λ)mink,mX (λ).

Because Yk = Yk+1

∣∣Nk(X,λ)

, again by exercise ?? we get

(γ) Mk divides Mk+1, for all k ≥ 1.

Suppose now k ≥ mX(λ). Combining (β) and (γ) yields Mk(t) = MmX (λ)(t) = (t − λ)mX (λ).

In particular, we get (Yk − λ)mX (λ) = 0, i.e. (X − λI)mX (λ)∣∣Nk(X,λ)

= 0, forcing the inclusion

Nk(X,λ) ⊂ NmX (λ)(X,λ). This then gives in fact the equality Nk(X,λ) = NmX (λ)(X,λ) =

N(X,λ).

The next step would be to prove that, if k ∈ 1, . . . ,mX(λ), then Nk−1(X,λ) ( Nk(X,λ).

By the Descent Lemma (exercise ??), it suffices to check this only for k = mX(λ). Suppose

NmX (λ)−1(X,λ) = N(X,λ). Using (β) this will force (X − λI)mX (λ)−1∣∣N(X,λ)

= 0. This is

however impossible, because the minimal polynomial of X∣∣N(X,λ)

is (t− λ)mX (λ).

Definition. Let V be a finite dimensional vector space, and let X : V → Vbe a linear map. For each λ ∈ Spec(X) we define the sequence of integers

∆X(λ) = [dim N1(X, λ),dim N2(X, λ), . . . ,dim NmX(λ)(X, λ)].

The family DX =(∆X(λ)

)λ∈Spec(X)

is called the spectral invariant of X.

Comments. It is understood that the spectral invariant ∆X has the followingdata built in it:

• The spectrum Spec(X) as the indexing set.• The spectral multiplicities. For each λ ∈ Λ, the spectral multiplictymX(λ) is the length of the sequence ∆X(λ).

So, given X, Y ∈ Lin(V ), the equality ∆X = ∆Y means:


(i) Spec(X) = Spec(Y );(ii) mX(λ) = mY (λ), for all λ ∈ Spec(X) = Spec(Y );(iii) dim Nk(X, λ) = dim Nk(Y, λ), for all λ ∈ Spec(X) = Spec(Y ) and allk with 1 ≤ k ≤ mX(λ) = mY (λ).

116 Prove that the spectral invariant is a similarity invariant.Hint: Assume Y = SXS−1. We know already that the spectra and the multiplicities are equal.

Prove that for any λ ∈ Spec(X) = Spec(Y ) and any integer k ≥ 1 we have S(Nk(X,λ)

)=

Nk(Y, λ), so (rememeber that linear isomorphisms preserve dimensions) we get dimNk(X,λ) =

dimNk(Y, λ).

117 Prove that the linear maps X, Y ∈ Lin(C4) defined in exercise ?? havedifferent spectral invariants. In particular X is not similar to Y .

The remainder of this section is devoted to the converse of exercise ??:

Theorem 1.5.1. (Spectral Invariant Theorem) If V is a finite dimensionalvector space, and if X, Y : V → V are linear maps with DX = ∆Y , then X issimilar to Y .

The proof of this result is quite ellaborate, so we will break it into several steps.The first result we are concerned with is a set of restrictions that occur on the

spectral invariant.

118 Let V be a finite dimensional vector space, and let X : V → V be a linearmap. Prove that, for each λ ∈ Spec(X), the sequence(

dim[Nk(X, λ)/Nk−1(X, λ)])mX(λ)

k=1

is non-decreasing.Hint: Use the Descent Lemma (exercise ??) for the linear map T = X − λI.

Comments. Since we have strict inclusions Nk−1(X, λ) ( Nk(X, λ), k =1, . . . ,mX(λ), by exercise ??, it follows that the terms in the above sequence alsosatsifies:

dim[Nk(X, λ)/Nk−1(X, λ)] ≥ 1, for all k = 1, . . . ,mX(λ).

We also know that, for each k ≥ 1, we have

dim[Nk(X, λ)/Nk−1(X, λ)] = dim Nk(X, λ)− dim Nk−1(X, λ).

So the spectral invariant exhibits the following pattern

dim N1(X, λ) ≥ dim N2(X, λ)− dim N1(X, λ) ≥ . . .

· · · ≥ dim NmX(λ)(X, λ)− dim NmX(λ)−1(X, λ) ≥ 1.(15)

119 Let V be a finite dimensional vector space, and let X : V → V bea linear map. Fix some λ ∈ Spec(X). Prove there exists a unique sequence[c1, c2, . . . , cmX(λ)] of non-negative integers, with cmX(λ) ≥ 1, such that for everyk = 1, . . . ,mX(λ), we have:

(16) dim Nk(X, λ) = c1 + 2c2c + · · ·+ kck + kck+1 + · · ·+ kcmX(λ).


Hint: Define first the numbers

pk = dim[Nk(X,λ)/Nk−1(X,λ)] = dimNk(X,λ)− dimNk−1(X,λ), k = 1, . . . ,mX(λ),

so that by exercies ?? we have

p1 ≥ p2 ≥ · · · ≥ pmX (λ) ≥ 1,

and

(17) dimNk(X,λ) = p1 + p2 + · · ·+ pk, for all k = 1, . . . ,mX(λ).

Define cmX (λ) = pmX (λ), and ck = pk − pk+1 for all k with 1 ≤ k < mX(λ). Then we will have

pk = ck + ck+1 + · · ·+ cmX (λ), for all k = 1, . . . ,mX(λ),

and the conclusion follows from (17)

Definitions. Let V be a finite dimensional vector space, and let X : V → Vbe a linear map. For every λ ∈ Spec(X), the numbers defined in the above exercisewill be called the λ-cell numbers of X. They will be denoted by ck(X, λ). From theabove discussion, it is clear that if one defines a system ΥX =

(ΥX(λ)

)λ∈Spec(X)

,where

ΥX(λ) = [c1(X, λ), c2(X, λ), . . . , cmX(λ)(X, λ)],then we have constructed a new similarity invariant, called the Jordan invariant.For each λ ∈ Spec(X), the set

bΥX(λ)c =k ∈ 1, . . . ,mX(λ) : ck(X, λ) 6= 0

will be called the λ-cell support of X.

Comment. It is clear that the spectral invariant and the Jordan invariant areequivalent, in the sense that

• ΥX = ΥX , if and only if ∆X = ∆Y .The proof of the Spectral Invariant Theorem will use the Jordan invariant in anessential way.

120 Let V be a finite dimensional vector space, and let T : V → V be a nilpotentlinear map, so that Spec(T ) = 0, and the minimal polynomial is MT (t) = tm forsome m ≥ 1. Prove that there eixsts a sequence of linear subspaces

Z1 ⊂ Nk(T, 0), Z2 ⊂ N2(T, 0), . . . , Zm ⊂ Nm(T, 0),

such that, for every p ∈ 1, . . . ,m we have:(i) Zp + T (Zp+1) + · · ·+ Tm−p(Zm) + Np−1(T, 0) = Np(T, 0);(ii) Zp ∩

[T (Zp+1) + · · ·+ Tm−p(Zm) + Np−1(T, 0)

]= 0.

Hint: Denote, for simplicity, the generalized eigenspaces Nk(T, 0) by Nk, k = 1, . . . ,m. Do

reverse induction on p. Start off by choosing a linear subspace Zm ⊂ Nm such that

• Zm +Nm−1 = Nm;

• Zm ∩Nm−1 = 0.The existence of such a subspace follows from exercise ??. Suppose the spaces Zk+1, Zk+2, . . . , Zm

have been chosen, such that properties (i) and (ii) hold for p ∈ k + 1, k + 2, . . . ,m. Considerthe linear subspace

Wk = T (Zk+1) + T 2(Zk+2) + · · ·+ Tm−k(Zm) +Nk−1 ⊂ Nk,

and use again exercise ?? to find a linear subspace Zk, such that

• Zk +Wk = Nk;• Zk ∩Wk = 0.


Definition. With the above notations, the sequence (Z1, Z2, . . . , Zm) will becalled a cell skeleton for T .

121 Use the setting and notations from exercise ??. Prove that if (Z1, . . . , Zm)is a cell skeleton for T , then for every p ∈ 1, . . . ,m we have the equality

Np(T, 0) =p∑

`=1

m∑j=`

T j−`(Zj).

(Here we use the convention: T 0(Zj) = Zj .)Hint: Use property (i) and induction on p.

In exercises ??-?? below, we fix a finite dimensional vector space V , a nilpotentlinear map T : V → V . Denote by m the degree of the minimal polynomial, i.e.we have MT (t) = tm. For simplicity, the generalized eigenspaces Nk(T, 0) will bedenoted by Nk, k = 1, . . . ,m.

122 Suppose p ∈ 1, . . . ,m. If W ⊂ Np is a linear subspace, with W ∩Np−1 =0. Suppose k is an integer with 1 ≤ k < p. Prove that

• The linear map T p−k∣∣W

: W → V is injective.• The space T p−k(W ) is a linear subspace of Nk, with T p−k(W ) ∩Nk−1 =0.

Hint: Notice that, by the construction of the generalized eigenspaces, as Nj = Ker(T j), we havethe inclusions

(18) T i(Nj) ⊂ Nj−i, for all i, j with 1 ≤ i < j ≤ m.

In particular, we have the inclusions

T p−k(Np) ⊂ Nk and T p−k(Np−1) ⊂ Nk−1.

From the first inclusion, we immediately get the inclusion T p−k(W ) ⊂ Nk. Using the FactorizationLemma (exercise ??), there exists a linear map S : Np/Np−1 → Nk/Nk−1, such that the diagram

Npquotient map−−−−−−−−−→ Np/Np−1

T p−k

y yS

Nkquotient map−−−−−−−−−→ Nk/Nk−1

is commutative. So, if we denote by qj : Nj → Nj/Nj−1, j = 1, . . . ,m, the quotient maps, wehave

(19) S qp = qk T p−k.

It turns out that S is injective. (Indeed, if we start with some element x ∈ Np/Np−1 with

S(x) = 0, then if we write x = qp(v) for some v ∈ Np, then we get 0 = S(x) = S(qp(x)

)=

qk(T p−k(v)

), which means that T p−k(v) belongs to Nk−1 = Ker(Tk−1). This will then give

T p−1(v) = Tk−1(T p−k(v)

)= 0, which means that v belongs to Ker(T p−1) = Np−1. Finally this

forces x = qp(v) = 0.)Now, since W ∩Ker qp = W ∩Np−1 = 0, by exercise ?? we know that the restriction qp

∣∣W

is injective. This gives the fact that the map (S qp)∣∣W

is injective.

When the equality (19) to W we see that the composition (qk T p−k)∣∣W

is injective, so this

will force T p−k∣∣W

to be injective. Moreover, the injectivity of (qk Tp− k)∣∣W

will also force the

equality T p−k(W ) ∩Ker qk = 0, which means precisely that T p−k(W ) ∩Nk−1 = 0.

123* Let (Z1, Z2, . . . , Zm) be a cell skeleton for T . Fix some integer p with1 ≤ p < m. Define the linear subspace

Vp = T (Zp+1) + T 2(Zp+2) + · · ·+ Tm−p(Zm).


Let Xp : Zp+1⊕Zp+2⊕ · · · ⊕Zm → Vp be the unique linear map with the propertythat Xp

∣∣Zp+k

= T k∣∣Zp+k

, for all k = 1, 2, . . . ,m− p. Prove that

(i) Xp is a linear isomorphism.(ii) Vp ∩Np−1 = 0.

Hint: Use reverse induction on p. Start with p = m − 1. In this case Vm−1 = T (Zm), andeverything is obvious from exercise ??. Assume now the properties (i) (ii) are true for p = k + 1,

and let us prove them for p = k. By construction, the linear subspace Zp+1 ⊂ Np+1 is constructed

in such a way that

(α) Zk+1 + Vk+1 +Np = Nk+1;

(β) Zk+1 ∩ [Vk+1 +Nk] = 0.Using the inductive hypothesis, we know that Vk+1 ∩ Nk = 0, which means that Vk+1 and

Nk form a linearly independent pair of subspaces. In particular, using exercise ?? and condition(β), we get the fact that the triple (Zk+1, Vk+1, Nk) is also a linearly independent family. In

particular, the subspace Yk+1 = Zk+1 + Vk+1 has the properties:

(γ) Yk+1 ∩Nk = 0;(δ) the linear map

Y : Zk+1 ⊕ Vk+1 3 (z, v) 7−→ z + v ∈ Yk+1

is a linear isomorphism.

Using (γ) and exercise ??, we know that T∣∣Yk+1

is injective, and T (Yk+1)∩Nk = 0, thus giving

property (i) for p = k. Observe that

T (Yk+1) = T (Zk+1) + T (Vk+1) = T (Zk+1) + T 2(Zk+2) + · · ·+ Tm−k(Zk) = Vk,

so we get a linear isomorphism T∣∣Yk+1

: Yk+1 → Vp. Using (δ), and the inductive hypothesis, we

immediately obtain property (ii) for p = k.

124 Let (Z1, Z2, . . . , Zm) be a cell skeleton for T . Prove the equalities:

dim Zp = cp(T, 0), for all p = 1, . . . ,m,

where c1(T, 0), . . . , cm(T, 0) are the cell numbers of T .Hint: Define, Vp = T (Zp+1) + · · ·+ Tm−p(Zm), and Wp = Vp +Np−1, for all p = 1, . . . ,m− 1.

Also put Wm = Zm. Denote dimZk simply by ck. On the one hand, by the preceding exercise,

we know that

(20) dimVp = dimZp+1 + dimZp+2 + · · ·+ dimZm = cp+1 + cp+2 + · · ·+ cm,

for all p = 1, . . . ,m− 1. On the other hand, again by the preceding exercise, we know that

(21) dimWp = dimVp + dimNp−1.

Recall that, by construction, the subspace Zp ⊂ Np is chosen such that

• Zp ∩Wp = 0;• Zp +Wp = Np.

This will give dimNp = dimZp + dimWp = cp + dimWp. Using (21) and (20) we get

dimNp = cp + dimVp + dimNp−1 = cp + cp+1 + · · ·+ cm + dimNp−1.

By the construction of the cell numbers (see exercise ??) this gives the desired result.

125 Let us denote by J the cell support of T , i.e.

J =j ∈ 1, . . . ,m : cj(X, 0) 6= 0

.

Let (Z1, . . . , Zm) be a cell skeleton for T . Prove that the family(T k(Zj)

)j∈J

0≤k<j

is a direct sum decomposition of V . (Here we use the convention: T 0(Zj) = Zj .)


Hint: On the one hand, we know that

dimTk(Zj) = dimZj = cj(X, 0), for all j, k with 0 ≤ k < j ≤ m.

Of course, if j 6∈ J , these dimensions are zero. In particular, using the properties of cell numbers

(exercise ??), we get we have

∑j∈J

0≤k<j

dimTk(Zj) =

m∑j=1

j−1∑k=0

dimTk(Zj) =

m∑j=1

j−1∑k=0

cj(X, 0) =

m∑j=1

jcj(X, 0) =

= dm(X, 0) = dimNm = dimV.

(22)

On the other hand, by exercise ??

V = Nm =

m∑`=1

m∑j=`

T j−`(Zj),

and after making the change of indices k = j − `, we get

V =

m∑j=1

j−1∑k=0

Tk(Zj) =∑j∈J

0≤k<j

Tk(Zj),

and the result then follows from (22) and exercise ??.

126* Prove the following

Theorem 1.5.2. (Spectral Invariant Theorem for Nilpotents) Let V and W befinite dimensional vector spaces, and let T : V → V and X : W → W be nilpotentlinear maps. Assume

dim[Ker(T k)] = dim[Ker(Xk)], for all k ≥ 1.

Then there exists a linear isomorphism S : V →W such that X = STS−1.

Sketch of proof: Denote for simplicity the numbers dim[Ker(Tk)] = dim[Ker(Xk)] by dk. We

know that we have 0 < d1 < · · · < dm = dm+1 = . . . , where m is the degree of the minimalpolynomial. In particular, we get:

• the minimal polynomials of T and X coincide: MT (t) = MX(t) = tm;

• dimV = dimW = dm.

As a consequence of the hypothesis, we also get the equality of the cell numbers:

ck(T, 0) = ck(X, 0), for all k ∈ 1, . . . ,m.

Denote, for simplicity these cell numbers by ck. Define the set

J =j ∈ 1, . . . ,m : cj 6= 0

.

Let (Z1, . . . , Zm) be a cell skeleton for T , and let (G1, . . . , Gm) be a cell skeleton for X. For each

j ∈ J , define the subspaces

Vj = Zj + T (Zj) + · · ·+ T j−1(Zj);(23)

Wj = Gj +X(Gj) + · · ·+Xj−1(Gj).(24)

Fix for the moment j ∈ J . We know that

(i) dimZj = dimGj = cj(> 0);

(ii) the families(Tk(Zj)

)j−1

k=0and

(Xk(Gj)

)j−1

k=0are linearly independent;

(iii) the linear maps Tk : Zj → Tk(Zj) and Xk : Gj → Xk(Gj), k = 1, . . . , j − 1 are linearisomorphisms.


Using (i), there exists a linear isomorphism Sj0 : Zj → Gj . Using (iii), for each k = 1, . . . , j − 1

there exist linear isomorphisms Sjk : Tk(Zj) → Xk(Gj) such that the diagram

ZjT k

−−−−−→ Tk(Zj)

Sj0

y ySjk

GjXk

−−−−−→ Xk(Zj)

is commutative. (Simply define Sjk = Xk Sj0 (Tk∣∣Zj

)−1.) By (ii), we know that (23) and (24)

are in fact direct sum decompositions. In particular, there exists a (unique) linear isomorphismSj : Vj →Wj such that

Sj

∣∣T k(Zj)

= Sjk, for all k = 0, . . . , j − 1.

It is not hard to prove that Vj is invariant for T , and Wj is invariant for X. Moreover, by

construction we will now have a commutative diagram:

(25)

VjT−−−−−→ Vj)

Sj

y ySj

WjX−−−−−→ Wj

Let now j vary in J . Using exercise ?? we have direct sum decompositions

V =∑j∈J

Vj and W =∑j∈J

Wj ,

so there exists a unique linear isomorphism S : V →W such that

S∣∣Vj

= Sj , for all j ∈ J.

Using (25) we then get ST = XS, and we are done.

127* Prove the Spectral Invariant Theorem.Hints: Assume X,Y : V → V have the same spectral invariants. Denote the set Spec(X) =

Spec(Y ) simply by Λ. We know that the minimal polynomials of X and Y coincide

MX(t) = MY (t) =∏λ∈Λ

(t− λ)m(λ).

Put Vλ = N(X,λ) and Wλ = N(Y, λ). Use the spectral decomposition, to write direct sumdecompositions

(26) V =∑λ∈Λ

Vλ =∑λ∈Λ

Wλ.

Fix for the moment λ ∈ Λ. We know that Vλ is invariant for X, and Wλ is invariant for Y .

Moreover, if we define Xλ = X∣∣Vλ

∈ Lin(Vλ) and Yλ = Y∣∣Wλ

∈ Lin(Wλ), then these linear maps

will have the same spectral invariants. We have Spec(Xλ) = Spec(Yλ) = λ, and

dk(Xλ, λ) = dk(X,λ) = dk(Y, λ) = dk(Yλ, λ).

It follows that the nilpotents Xλ − λI and Yλ − λI have the same spectral invariants. By the

Spectral Theorem for Nilpotents, there exists a linear isomorphism Sλ : Vλ → Wλ such thatSλ(Xλ − λI) = (Yλ − λI)Sλ. This clearly gives

(27) SλXλ = YλSλ.

Let now λ vary. Using the direct sum decompositions (26), there exists a (unique) linear isomor-phism S : V → V , such that

S∣∣Vλ

= Sλ, for all λ ∈ Λ.

Using (27), we immediatel get SX = Y S, and we are done.

CHAPTER 2

Matrices

1. Arithmetic of matrices

Definitions. Suppose k is a field, and m,n ≥ 1 are integers. A rectangulartable:

(28)

a11 a12 . . . a1n

a21 a22 . . . a2n

...... . . .

...am1 am2 . . . amn

,

filled with elements of k, is called an m× n matrix with coefficients in k. The setof all such matrices will be denoted by Matm×n(k).

A square matrix is one with m = n. For the set of all square n × n matrices,we will use the simplified notation Matn(k).

The indexing convention for writing the coefficients will be that the first indexrepresents the row number, and the second index represents the column number.So the matrix (28) can also be written as[

aij

]1≤i≤m1≤j≤n

.

128 Prove that Matm×n(k) is a k-vector space, when equipped with the followingoperations:

a11 a12 . . . a1n

a21 a22 . . . a2n

...... . . .

...am1 am2 . . . amn

+

b11 b12 . . . b1n

b21 b22 . . . b2n

...... . . .

...bm1 bm2 . . . bmn

=

=

a11 + b11 a12 + b12 . . . a1n + b1n

a21 + b21 a22 + b22 . . . a2n + b2n

...... . . .

...am1 + bm1 am2 + bm2 . . . amn + bmn

,

λ ·

a11 a12 . . . a1n

a21 a22 . . . a2n

...... . . .

...am1 am2 . . . amn

=

λa11 λa12 . . . λa1n

λa21 λa22 . . . λa2n

...... . . .

...λam1 λam2 . . . λamn

The zero matrix in 0m×n ∈Matm×n(k) is the matrix filled with zeros. Prove thatdim Matm×n(k) = mn.

39

40 2. MATRICES

Matrices arise naturally in the study of linear maps between finite dimensionalvector spaces.

Definition. Suppose m,n ≥ 1 are integers. Let V be a vector space of di-mension n, and let W be a vector space of dimension m. Suppose we have twoordered1 linear bases A = (v1, . . . , vn) for V , and B = (w1, . . . , wm) for W . Sup-pose a linear map T : V → W is given. For each j ∈ 1, . . . , n, there are uniquescalars α1j , α2j , . . . , αmj such that

T (vj) = α1jw1 + α2jw2 + · · ·+ αmjwm.

The m× n matrix

MA,BT =

α11 α12 . . . α1n

α21 α22 . . . α2n

...... . . .

...αm1 αm2 . . . αmn

is called the matrix of T , with respect to the pair (A,B). The map

ΩAB : Lin(V,W ) 3 T 7−→Matm×n(k)

is called the matrix representation associated with the pair (A,B).

129 With the above notations, prove that the matrix representation

ΩAB : Lin(V,W )→Matm×n(k)

is a linear isomorphism.

Example. The simplest example of the above construction is the following.Assume V = k, with ordered basis A = (1), and W = k, with the standard orderedlinear basis B = (e1, . . . ,em). We have the following bijective correspondences:

Lin(k,km) 3 T 7−→ MABT ∈Matm×1(k)(29)

Lin(k,km) 3 T 7−→ T (1) ∈ km.(30)

Composing the correspondence (29) with the inverse of the correspondence (30),we get a bijective correspondence km →Matm×1(k).

130 Prove that the bijective correspondence described above is given as

km 3 (α1, . . . , αm) 7−→

α1

α2

...αm

.

Convention. From now on we will think km as identified with the collectionof m× 1 matrices. Such matrices are called column matrices.

1 An ordered linear basis is a prefered listing of a linear basis (by means of specifying whichis the first vector, the second vector, etc.)

1. ARITHMETIC OF MATRICES 41

Comment. With the above identification, the standard ordered linear basis(e1, . . . ,em) of km is described by

ei =

0...1...0

← i , i = 1, . . . ,m.

131 Let m,n ≥ 1 be integers, and let T : kn → km be a linear map. Let A =(e(n)

1 , . . . ,e(n)n ) be the standard orderd basis for kn, and let B = (e(m)

1 , . . . ,e(m)n )

be the standard orderd basis for km. Prove that the matrix of T with respect to(A,B) is given by

MABT =

[T (e(n)

1 )∣∣∣∣T (e(n)

2 )∣∣∣∣ . . .

∣∣∣∣T (e(n)n )

].

That is, the columns of MABT are precisely the column matrices T (e(n)

1 ), . . . T (e(n)n ).

Definition. Suppose m,n, p ≥ 1 are integers, and we have matrices X ∈Matm×n(k), and Y ∈Matn×p(k), say

X =[xij

]1≤i≤m1≤j≤n

and Y =[yk`

]1≤k≤n1≤`≤p

.

Construct the matrixZ =

[zij

]1≤i≤m1≤j≤p

∈Matm×p(k)

by defining

zij = xi1y1j + xi2y2j + · · ·+ xinynj , for all i ∈ 1, . . . ,m, j ∈ 1, . . . , p.

The matrix Z is called the product of X and Y , and is simply denoted by XY .

The following exercise explains why the product is defined in this way.

132 Let m,n, p ≥ 1 be integers, and let U, V,W be vector spaces with dim U = p,dim V = n and dim W = m. Fix an ordered linear bases A for U , B for V , andC for W . Suppose we are given linear maps S : U → V and T : V → W . LetX ∈Matm×n(k) be the matrix of T with respect to (B,C), and let Y ∈Matn×p(k)be the matrix of S with respect to (A,B). Prove that the matrix of T S : U →W ,with respect to (A,C) is equal to the product XY .

Comments. The above result says that we have a commutative diagram

(31)

Lin(V,W )× Lin(U, V )composition−−−−−−−→ Lin(U,W )

ΩBC×ΩAB

y yΩAC

Matm×n(k)×Matn×p(k) −−−−−→product

Matm×p(k)

133 Use (31) to derive the following properties of the product:

42 2. MATRICES

A. The product is associative. That is, given matrices X ∈ Matm×n(k),Y ∈Matn×p(k), and Z ∈Matp×r(k), one has the equality

(XY )Z = X(Y Z).

B. Given matrices M ∈Matm×n(k) and P ∈Matp×r(k), the maps

Matn×p(k) 3 X 7−→MX ∈Matm×p(k)

Matn×p(k) 3 X 7−→ XP ∈Matn×r(k)

are linear

Definition. Given an integer n ≥ 1, we define the identity matrix

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

134 Let V be a vector space of dimension n, and let B an ordered linear basis

for V . Prove that ΩBB(IdV ) = In.In particular, using (31), obtain the identities

ImX = XIn = X, for all X ∈Matm×n(k).

Notation. Let n ≥ 1 be an integer, let V be a vector space of dimension n, andlet B be an ordered linear basis for V . The linear map ΩBB : Lin(V )→Matn(k)will be denoted by ΩB .

Comments. The above results, when applied to square matrices, give the factthat Matn(k) is a unital k-algebra. In particular (see Section 3 in Chapter 1), forany matrix M ∈Matn(k) one has a polynomial functional calculus

k[t] 3 P 7−→ P (M) ∈Matn(k).

If V is a vector space of dimension n, and if B is an ordered linear basis for V ,then

ΩB : Lin(V )→Matn(k)

is an isomorphism of k-algebra. It is clear that, for a linear map T : V → V , wehave

ΩB

(P (T )

)= P

(ΩB(T )

), for all P ∈ k[t].

135 Let n ≥ 1 be an integer, and let T : kn → kn be a linear map. Let A

136 Let m,n ≥ 1 be integers, and let T : kn → km be a linear map. LetM ∈ Matm×n(k) be the matrix of T with respect to the standard ordered linearbases. Prove that

(32) T (X) = MX, for all X ∈ kn.

(Recall that we identify kn with Matn×1(k).)

2. GAUSS-JORDAN ELIMINATION 43

Definition. Let m,n ≥ 1 be integers, and let T : kn → km be a linear map.An m× n matrix M , satisfying (32) is said to represent T .

137 Let m,n ≥ 1 be integers, and let T : kn → km be a linear map. Prove thatthe matrix M connstructed in exercise ?? is the unique one that represents T .

138 Let n ≥ 1 be an integer, and let X be an n × n matrix. Prove that thereexists a unique non-zero polynomial M ∈ k[t] with the following with the followingproperties:

(i) M(X) = 0.(ii) M is monic, in the sense that the leading coefficient is 1.(iii) If P ∈ k[t] is any polynomial with P (X) = 0, then P is divisible by M

Hint: Use the same arguments as in exercise ??.

Definition. The polynomial described above is called the minimal polynomialof the matrix X, and is denoted by MX .

139 Let n ≥ 1 be an integer, let V be a vector space of dimension n, and let Bbe an ordered linear basis for V . Suppose a linear map T : V → V is given. Definethe n × n matrix X = ΩB(T ). Prove that the minimal polynomials of T and Xcoincide:

MX(t) = MT (t).

Hint: Use the fact that ΩB is an isomorphism of k-algebras.

140 Let n ≥ 1 be an integer, and let A ∈Matn(k). Define the linear map

TA : kn 3 X 7−→ AX ∈ kn,

represented by A. Prove that the minimal polynomials of T and A coincide:

MT (t) = MA(t).

Hint: Show that the correspondence

Matn(k) 3 A 7−→ TA ∈ Lin(kn)

is an isomorphism of k-algebras.

Using this fact, one can derive the The matrix representations of linear mapsThe matrix operationsRankSystems of linear equations

2. Gauss-Jordan elimination

Row transformationsGauss-Jordan reductioninvariance of rank; rank test for consistencyApplications to finding the kernelAplications to completing basis

44 2. MATRICES

3. Similarity

The similarity problemUpper triangular forms

4. Determinants

Def. of determinantdet(AB)=det(A)det(B)row/column expansionlinear independence via determinantsinverse of a matrix via determinants

5. A conceptual approach to determinants*

Tensor productTensor algebraExterior algebraMeaning of determinant and its basic properties

CHAPTER 3

The Jordan canonical form

1. Characteristic polynomial

Def. of char polynRoots of char polyn = eigenvaluesCayley-Hamilton thmMultiplicity in char polyn = dim of spectral subspaces

2. Jordan basis

The algorithmThe Jordan canonical formSpectral multiplicity diagram vs. Jordan canonical formConclusion: spectral multiplicity diagram is a complete invariant for the simi-

larity problemJordan dec: SS+NILP

3. Applications

Entire functionsExponentials and trigonometric functionsHolomorphic functional calculus*Spectral mapping thm

4. Real Jordan form*

ComplexificationMinimal/char real polynomial, vs. minimal/char complex polynomialReal Jordan formApplications

45

CHAPTER 4

Linear algebra on finite dimensional Hilbert spaces

1. Hilbert spaces

Inner products and normsCBS ineq.Orthon. basis

2. Linear maps on Hilbert spaces

The adjointDuality ker X*, Ran X*, etc.The norm of a linear mapSpectral properties of the adjoint

3. Normal operators

Normal and self-adjoint mapsPositivityThe unitary groupPolar decomposition*Generalized eig spaces for normal operators = eigensppaceThe spectral thm

4. Applications

Spectral thm for self-adjointFunctional calculus depends only on the values on the spectrumSquare rootsPositivity via diagonal minors

47

Appendices

A. Results from group theory

B. The symmetric group

49

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BASIC LINEAR ALGEBRA - Kansas State Universitynagy/lin-alg/notes.pdfCHAPTER 1 Vector spaces and linear maps In this chapter we introduce the basic algebraic notions of vector spaces